Part 1: The Algebra-based Approach
Simple Harmonic Motion (SHM):
A mass attached to a linear spring and set into up-and-down motion performs a motion that is called the "oscillatory motion", "simple harmonic motion, or SHM." We need to first study the behavior of a linear spring.
In the 3 cases shown in Fig. 1, the top one is the un-stretched spring that is fixed at its left end. When force F1 pulls it to the right its length becomes X1. When a greater force F2pulls it to the right its length becomes X2. The change in force is ΔF = F2 - F1, and the change in length is ΔX = X2 - X1. Now, if we double ΔF and observe that ΔX also doubles, or if we triple ΔF and observe that ΔX also triples, we conclude that the spring is a linear one. Such spring behavior leads us to the following definition:
A linear spring is one for which the change in length Δx is proportional to the change in the applied force ΔF.
ΔF = k Δx
where k is called the "spring constant." The SI unit for k is N/mand the American unit is lb/ft.
Example 1: A linear spring has a free length of 18.0cm. When it is under a hanging load of 125N, its total length is 20.5cm. Calculate (a) its constant k and (b) the hanging load that makes it 25.0cm long. For convenience, let the +y axis be downward.
Solution: (a) ΔF
= kΔy ;
125N - 0 = k (20.5-18.0)cm ; k = 50.0 N/cm.
(b) ΔF = k Δy ; F2 - 0 = 50.0 N/cm (25.0 - 18.0)cm ; F2 = 350N.
Example 2: A linear spring has a length of 35.0cm when under a hanging load of 225N and a length of 43.0cm when under a hanging load of 545N. Find (a) its constant, and (b) its free (no load) length.
Solution: To be solved by students.
The Formula for Linear Springs: shows a no load spring. If it is pulled to the right with Fappl. as in Fig. 3a the spring pulls to the left with force Fs. Now if the spring is pushed to the left as in Fig. 3c with Fappl., it pushes to the right with force Fs. Since, as we already know Fappl. = kx ; therefore, Fs = -kx.
The Linear Spring Formula:
Fs = -kx where Fs is the spring force.
Simple Harmonic Motion (SHM):
Recall the definition of angular speed ω: ω = Δθ/Δt.
For constant ω, θ is a linear function of t. We may write: ω = θ/t or
θ = ωt. This will be used later.
In Fig. 4., with parallel rays of light casting down, as Particle M performs uniform circular motion, the shadow of M on the x-axis performs repeated back-and-forth motion that is called the "simple harmonic motion" or the "oscillatory motion." θ determines the angular position of M on the circle at time t while K' determines the horizontal position or distance of the shadow from O' at time t. K' or the shadow of M keeps moving back and forth between -A and +A as M rotates on the circle. Referring to the derivation on the diagram, we may write: cosθ = X/A or X = Acosθ or
|The Equation of the "Oscillatory Motion" of shadow K':||
x = Acos(ωt)
Read carefully: In its rotation, as M goes from P1 to P2, (K')'s distance to O' or X decreases from A to 0. In the 2nd Quadrant, as Particle M goes from P2 to P3, K' goes from O' to -A that means X changes from 0 to -A. In the 3rd Quadrant, as M goes from P3 to P4, K' returns from -A back to O' or X goes from -A to 0. Finally, when M goes from P4 to P1, K' goes from O' to +A and the cycle completes. One full turn of M from P1 and back to P1 results in one full oscillation of K' from +A to -A and back to +A.
The graph of x versus θ or ωt is shown in Fig. 5. The graph is based on the assumption that at t = 0, θ = 0. That is, Particle M starts its motion from P1. Note that the farthest distance Point K' can go from Point O' is +A or -A , or simply |A|, the radius of the circle. A that is the maximum deviation from the equilibrium position (O') is called the "Amplitude" of oscillations.
The equilibrium position for oscillator K' is Point O'. Again, do not forget that our goal is to study the motion (oscillation) of shadow K' on the x-axis.
Example 3: A bicycle wheel of radius 30.0cm is spinning at a constant angular speed of 180 rpm in a vertical plane. Find (a) its angular speed in rd/s. The shadow of a bump on its edge performs an oscillatory motion on the floor. Write (b) the equation of the oscillations of the shadow knowing that the shadow is at its maximum at t = 0. (c) determine the distance of the shadow from the equilibrium position at t = 1.77s.
Solution: (a) ω = 180 (rev/min) = 180(6.28rd /60s) = 18.8 rd /s.
b) The constants are: ω = 18.8rd/s & A = 30.0cm ; the variables are x and t.
The general form is: X = Acos(ωt). Here: x = [30.0cm]cos (18.8t ).
(c) For t = 1.77s, x becomes: x = (30.0cm)cos(18.8*1.77rd) = -8.56cm.
Note that your calculator must be in radians mode for the last calculation.
Example 4: The equation of oscillations of a mass on a spring is given by x = 3.23 cos(12.56t) where x is in (cm) and t in seconds. Find its (a) amplitude, (b) angular speed, (c) frequency and period of oscillations, and (d) its position at t = 0.112s.
Solution: Comparing the given equation with the general form x = Acos(ωt), it is clear that
(a) A = 3.23cm ; (b) ω = 12.56 rd/s ; (c) ω =2πf ; f = ω/2π ; f = 2.00Hz.
T = 1/f ; T = 0.500s ; (d) x = 3.23 cos(12.56*0.112 rd) = 0.528cm.
Note that your calculator must be in radians mode for the last calculation.
Another Equation Form for SHM:
Parallel rays of light going to the left cast the shadow of the circulating Particle M on the left wall as shown in Fig. 6. As M rotates at constant ω, Shadow K" performs repeating up and down motion or oscillations on the wall. Again, if exactly when M is passing by point P1, a stop watch is turned on (t = 0), at t = 0, Shadow K" starts from O" going upward. When M goes from P1 to P2 on the circle, its shadow on the wall goes from O" to +A or y varies from 0 to +A.
When M goes from P2 to P3, Shadow K" goes down from +A to O" or y changes from +A to 0. As M goes from P3 to P4, K" goes from O" to -A or y changes from 0 to -A. Finally when M goes from P4back to P1, K" goes from -A to O", or y changes from -A to 0 and completes the cycle. One full turn of M from P1 and back to P1 results in one full oscillation of K" from 0 to +A, back to 0, further to -A, and back to 0.
The graph of y versus θ or ωt is shown in Fig. 7. The graph is based on the assumption that at t = 0, θ = 0. That is, Particle M starts its motion from P1. Note that the farthest distance Point K" can go from Point O" is +A or -A, or simply |A|, the radius of the circle. A that is the maximum deviation from the equilibrium position (O") is called the "Amplitude" of oscillations. The equilibrium position for oscillator K" is Point O". Again, do not forget that our goal is to study the motion (oscillations) of shadow K" on the y-axis.
Example 5: A student in the passenger seat of a car traveling at night notices the oscillating reflectors on the heals of a biker that is pedaling far away. Although the motion of the heal of the biker is circular; however, the motion of each heal appears oscillatory from far away. He turns a stop watch on and measures the time of 100 full oscillations of one reflector to be 40 seconds. He estimates the distance from highest to lowest points to be 44cm. If the instant the stop watch is turned on is when the pedal is just passing the midpoint in its upward motion, find the (a) amplitude, (b) period, (c) frequency, (d) angular frequency of the oscillations and (e) write the equation of motion for such oscillations. Suppose the biker has a constant velocity forward. That way, his pedaling is at constant angular speed.
Solution: (a) A = 44cm/2 = 22cm ; (b) T = 40s/100 = 0.40s ;
(c) f = 1/T = 1/0.40s = 2.5/s ; (d) ω = 2πf = 2π(2.5/s) = 16 rd/s ;
(e) Using A=22cm & ω = 16 rd/s y = [22cm] sin(16t).
Note: Since at t = 0, y = 0 and y starts to increase to +A, the equation of oscillations has a sine form.
The Mass-Spring System:
When mass M hung from a linear spring is pulled down and released, its up-and-down motion above and below the equilibrium level is called the "simple harmonic motion." In the absence of frictional forces, the graph of such motion as a function of time has a perfect "sine" shape. It is for this reason that the motion is called harmonic. Fig. 8a shows an unloaded (unstretched)spring. Fig. 8b shows the same spring but loaded and therefore stretched as much as (-h). Fig. 8c shows that the loaded spring is further pulled down to (-A) and released. It shows that M oscillates up and down to +A and -Aabove and below the equilibrium level (the dotted line).
Using higher level mathematics, it can be shown that the angular frequency of oscillations of a mass-spring system is given by
From ω we may then calculate period T and frequency f. Note that the lower end of the spring is used as the "equilibrium level or position" for our study of the oscillations. We could have chosen any other level and kept it consistent in all figures.
Example 6: A 102-gram mass hung from a weak spring stretches it by 3.00cm. Let g = 9.81m/s2 and calculate (a) the load on the spring and (b) the spring constant in N/m. If this mass-spring system is initially in static equilibrium and at rest, and the mass is pushed up by +2.00cm and released, calculate its (c) angular speed, (d) frequency, (e) period, (f) the amplitude of oscillations, and (g) the equation of motion of such oscillations.
Solution: (a) The load is w = Mg ; w = (0.102kg)(9.81 m/s2) = 1.00N.
(b) ΔF = k Δx ; k = (1.00N) /(0.0300m) = 33.3 N/m.
(c) ω = (k/M)1/2 = [(33.3 N/m)/(0.102 kg)]1/2 = 18.1 rd/s.
(d) f = ω/(2π) ; f = 2.88 Hz.
(e) T = 1/f ; T = 0.347s.
(f) The 2.00cm that M is initially pushed up above its equilibrium level, becomes the amplitude of oscillations. A = +2.00cm.
(g) Knowing the constants: A = 2.00cm & ω = 18.1 rd/s, the equation of motion becomes:
x = [2.00cm] cos (18.1t ).
In this equation, if we plug t = 0, we get X = +2.00cm. This is correct because at t = 0, the mass is pushed up to X = +2.00cm and released.
How To Determine The Type of Sinusoidal Equation?
Suppose you have a stop watch in one hand and with the other hand you can change the position of a mass that is hanging from a linear spring and put it into oscillation. Visualize a mass hanging from a linear spring that is initially at rest. Suppose you do the following four experiments:
1) You push the mass slightly up to +A and release it exactly at the instant you turn the stop watch on (t = 0). Since at t = 0, the mass is at its maximum position (+A), the equation of its oscillations will have a cosine form:
X = +Acos(ωt) as shown in the left figure below.
2) You pull the mass slightly down to -A and release it exactly at the instant you turn the stop watch on (t = 0). Since at t = 0, the mass is at its lowest position (-A), the equation of its oscillations will have a (-) cosine form: X = -Acos(ωt) as shown in Fig. 9.
3) You strike the mass from underneath in the upward direction at the instant you turn the stop watch on (t = 0). Since at t = 0, the mass is at its equilibrium position, x = 0, with an upward initial speed enough to reach the intended (+A), its equation of motion has a sine form: X = +Asin(ωt) as shown in the left figure below.
4) You strike the mass from its top in the downward direction at the instant you turn the stop watch on (t = 0). Since at t = 0, the mass is at its equilibrium position, x = 0, with a downward initial speed enough to reach the intended (-A), the equation of its oscillations has a (-) sine form: X = -Asin(ωt) as shown in Fig. 10.
5) If you either pull the mass down or push it up, but not release it, instead give it an initial speed in either downward or upward direction, its equation of oscillations will not have any of the above four forms. It will have the general form below with a phase angle, φ, that in general is neither 0 nor π/2. We do not go into more details on this topic here.
X = Asin (ωt + φ )
Example 7: The graph of x (the distance from the equilibrium position) versus time (t) for the oscillations of a mass-spring system is given in Fig. 11.
For such oscillations, find (a) the amplitude, (b) the period, (c) the frequency, (d) the angular frequency, (e) the spring constant, k, if the mass of the object is 250 grams, and (f) the equation of motion for such oscillations.
Solution: (a) A = 2.00cm ; (b) T = 2(0.125s) = 0.250s
(c) f = 1/T ; f = 4.00 Hz ; (d) ω = 2πf ; ω = 2π(4.00/s) = 25.1 rd/s.
(e) ω = (k/M)(1/2) ; ω2 =(k/M); k = Mω2; k =(0.250kg)(25.1rd/s)2; k = 158N/m.
(f) x = Asin(ωt) ; x = [2.00cm]sin (25.12t). From the graph, the given graph is a sine function.
Linear Velocity and Acceleration in Simple Harmonic Motion:
Velocity: If an object is oscillating back and forth on a frictionless surface as shown in Fig. 12, it is easy to see that its velocity becomes zero at the extreme ends (a) and (c). This is simply because it has to first come to stop at the end points before it can return. It is also easy to see that velocity gains its maximum magnitude at the midpoint or the equilibrium position (b). We may therefore state that:
"In SHM, maximum speed occurs at x = 0 (the equilibrium position), and zero speed occurs at end points (x = +A or x = -A)."
Acceleration: As shown above, at the middle (x = 0), a = 0 because the spring is neither stretched nor compressed, Fs = 0. At end points, when the spring is at its maximum stretch or compress, the spring force is at its maximum magnitude, and therefore the acceleration that it gives to the attached mass is maximum in magnitude as well. We may therefore state that:
"In SHM, |amax.| occurs at end points where the force |F| is maximum, and a = 0 occurs at x = 0, the midpoint, where F = 0."
Using Calculus, if x = Acos(wt), the equations for x, v, and a in SHM are:
xmax., vmax. , and amax. are determined by setting each of cos(ωt) and sin(ωt) above equal to its maximum value that is 1. xmax, vmax , and amax become:
xmax = A , vmax. = - Aω and amax. = -Aω2.
We may ignore the (-) signs if only the magnitudes are concerned, as shown above. Note that xmax = A = amplitude.
Example 8: The equation of motion of a 22-kg log oscillating on ocean surface is x = 1.2 sin (3.14t) where x is in meters and t in seconds. Determine its, amplitude, angular speed, frequency, period, maximum speed, maximum acceleration (magnitude), and its position at t = 0.19s.
Solution: A =1.2m ; ω = 3.14rd/s ; f = ω/(2π) = 0.50/s ; T =1/f =2.0s.
|Vmax | = Aω ; |Vmax| = 1.2m(3.14 rd/s) = 3.8 m/s (occurs at the middle).
|amax | = Aω2 ; |amax| = (1.2m)(3.14rd/s)2 = 11.8 m/s2.
Using the given equation, substituting for t, and putting the calculator in "Radians Mode," we get:
x = 1.2m sin [3.14(0.19)rd] = 0.67m.
Energy Stored in a Linear Spring:
As a spring is stretched or compressed further and further, it stores more and more energy. To calculate the energy stored in a linear spring, we need to calculate the work done on it by the pushing or the pulling force. The work done on a spring equals the energy stored in it. This is simply the energy conservation law.
Work calculation is easy! Just multiply force F by displacement Δx. Of course, when we did that in Chapter 6, F was usually a constant force.
For a spring, F varies with x. Fortunately, for a linear spring, F varies linearly with x.
When you stretch a spring beyond its free length, the change in length goes from 0 to x as you change the applied force from 0 to F. The average force is (0+F)/2.
To calculate the work W or the energy stored U, just multiply F/2 by x.
W = (F/2)x or U = (1/2)Fx. Since for a linear spring, F = kx, W or U becomes: U = (1/2) (kx)x or
|Energy stored in a linear spring: U = (1/2)kx2.|
Example 9: Calculate the energy stored in linear spring that is compressed by 20.0cm under a 300kg load.
Solution: First, we need to find the spring constant k.
F = kx ; 300kg(9.81m/s2) = k (0.200m) ; k = 14700N/m.
U = (1/2)kx2 ; U = 0.5(14700N/m)(0.200m)2 ; U = 294J.
Test Yourself 1: For answers, click here.
1) A linear spring is one for which the relation between the applied force and length change is (a) F = k(Δx)2 (b) F2 = k(Δx) (c) F = k(Δx).
2) The relation between the applied force Fappl. to a spring and the change in its length Δx is (a) Fappl. = k(Δx) (b) Fappl. = - k(Δx) (c) neither a, nor b.
3) The relation between the spring force Fs and Δx is (a) Fs = k(Δx) (b) - Fs = - k(Δx) (c) Fs = - k(Δx).
4) The spring formula Fs = - k(Δx) is normally written as Fs = - kx (a) True (b) False click here.
5) If a spring stretches 5.00cm under a load of 100N, it has a constant of (a) 20.0N/cm (b) 2000N/m (c) both a & b.
6) A spring that stretches 7.05cm under a hanging 21.58-kg load has a constant of 20.N/cm (b) 15N/m (c) 30N/cm.
Problem: For the following questions, refer to Figure 4 and suppose that the radius of the circle is A = 10.0cm and mass M makes 5.00 rotations per minute: click here.
7) When θ = 30.0o, the distance from the position of the shadow, K', to the equilibrium position, O' is (a) 5.00cm (b) 8.66cm (c) 7.5cm. click here.
8) When θ = 60.0o, the distance from the position of the shadow, K', to the equilibrium position, O' is (a) 8.00cm (b) 7.66cm (c) 5.00cm.
9) When θ = 90.0o, the distance from the position of the shadow, K', to the equilibrium position, O' is (a) 0 (b) 5.00cm (c) 7.5cm.
10) When θ = 00.0o, the distance from the position of the shadow, K', to the equilibrium position, O' is (a) 10.0cm (b) 8.66cm (c) 7.5cm. click here.
11) When θ = 135.0o, the distance from the position of the shadow, K', to the equilibrium position, O' is (a) -10.0cm (b) 8.66cm (c) -7.07cm.
12) When θ = 225.0o, the distance from the position of the shadow, K', to the equilibrium position, O' is (a) -7.07cm (b) 8.66cm (c) -10.0cm.
13) When θ = 330.0o, the distance from the position of the shadow, K', to the equilibrium position, O' is (a) -7.07cm (b) 8.66cm (c) -10.0cm. click here.
14) Reading the problem's statement again, the angular speed ω of mass M is (a) 5.00rpm (b) 0.523 rd/s (c) both a & b.
15) The amplitude of oscillations is (a) 8.66cm (b) 0.707cm (c) 10.0cm. click here.
16) On the floor, the equation of oscillations of the shadow is (a) x = [10.0cm]cos(0.523t) (b) x = (1/2)(0.523 t2) (c) neither a nor b.
17) Setting t = 0 in this equation, gives us the position of the shadow at t = 0. Doing this, we get (a) x = 0.0 (b) x = 10.0cm (c) x = 8.66cm. click here.
18) Setting t = 1.10s in this equation, gives us the position of the shadow at t = 1.10s. Doing this, we get (a) x = 0.0 (b) x = 10.0cm (c) x = 8.39cm. [Is your calculator in radians mode? Incos(0.523*1.1), the angular speed 0.523 is in rd/s.]
19) Setting t = 2.00s in this equation, gives us the position of the shadow at t = 2.00s. Doing this, we get (a) x=5.01cm (b) x = 10.0cm (c) x = 0. click here.
20) Since ω = 0.523rd/s, and ω = 2πf , the value of f, (number of turns per second), is (a) f = 0.0833 s-1 (b) f = 0.0833/s (c) f = 0.0833 Hz (d) a, b, & c.
21) Based on the value of f, the value of period, T is (a) 12.0s (b) 12.0min (c) 1.2s. click here.
22) We could have also found the period, T from the information in the problem's statement that says 5.00 turns per minute. If in every minute 5.00 full turns are completed, then the time for completion of each turn T is (a) 12.0s (b) 2.00s (c) 8.00s.
Problem: The equation of motion for the oscillations of a mass attached to a spring is x = [4.00]cos(18.78t) where x is in cm and t is in seconds. Based on this equation,
23) the amplitude of oscillations is (a) 8.00cm (b) 4.00cm (c) 2.00cm.
24) the angular frequency ω or the # of radians swept per second is (a) 18.78 rd/min (b) 18.78 rd/s (c) 18.78o per second. click here.
25) The frequency of oscillations, f (the # of turns per second or the # of full oscillations per second) is (a) 3.00/s (b) 3.00s-1 (c) 3.00Hz (d) a, b, & c.
26) The period of oscillations T is (a) (1/3)s (b) (1/3)min (c) (1/3)yr. click here.
27) The object's distance to its equilibrium position at t = 0 is (a) 2.00cm (b) 6.00cm (c) 4.00cm.
28) The object's distance to its equilibrium position at t = 0.0175s is (a) 3.786cm (b) 3.21cm (c) 4.00cm.
29) The object's distance to its equilibrium position at t = 2.945s is (a) 2.28cm. (b) 1.293cm. (c) 1.82cm.
Problem: Refer to the figure of Example 7. There are 12 segments (time intervals) on the t-axis. Use the equation of motion found under (f) to calculate the following: Compare your calculations with the vertical line segments (x values) to see if they make sense. Make sure to perform all calculations with your calculator in the correct mode. click here.
30) The time interval corresponding to each segment is (a) 0.04167s (b) 0.02083s (c) 0.01042s.
31) Each vertical line under the sine-curve shows the distance of the object from (a) the equilibrium position (b) the mid-point (c) both a & b. click here.
32) The maximum distance you may calculate for vertical segments is (a) 1.00cm (b) infinity (c) 2.00cm.
33) The distance from Midpoint at t = 1(0.02083s) is (a) 2.00cm (b) 1.00cm (c) 3.00cm.
34) The distance from Midpoint at t = 2(0.02083s) is (a) 1.20cm (b) 1.73cm (c) 0. click here.
35) The distance from Midpoint at t = 3(0.02083s) is (a) 0.80cm (b) 2.00cm (c) 0.
36) The distance from Midpoint at t = 4(0.02083s) is (a) 0.60cm (b) 1.73cm (c) 0.50.
37) The distance from Midpoint at t = 5(0.02083s) is (a) 0.70cm (b) 1.00cm (c) 0.90. click here.
38) The maximum linear speed of a particle in oscillatory motion is (a) V = Aω (b) V = Aωt (c) V = Aωt2.
39) A, the amplitude, in oscillatory motion is the same thing as R, the rotation radius in circular motion. (a) True (b) False
40) In Example 7, the maximum linear speed of the mass is (a) 25.1 m/s (b) 50.2 rd/s (c) 50.2 cm/s.
41) Maximum linear speed of an oscillating mass occurs at (a) +A (b) -A (c) 0, the midpoint. click here.
42) In Example 7, the |amax.|of the mass is (a) Aω2 (b) Aαt (c) Aαt2.
43) Maximum linear acceleration (magnitude) of an oscillating mass occurs at (a) +A (b) -A (c) 0, the midpoint. (d) both a & b.
44) In Oscillatory motion, max. acceleration magnitude occurs at +/-A. The reason is that (a) the velocity at end points may not be zero (b) the force at end points has maximum magnitude and causes maximum acceleration (c) At end points, the spring is at maximum compress or maximum stretch (d) b & c. click here.
Problems Set 1:
1) Calculate (a) the change in length of a linear spring with a constant of 730N/m when under a load of 146N, and (b) 292N.
2) A 200.0-gram mass hanging from a linear spring gives it a length of 14.0cm. When the hanging mass is increased to 350.0 grams, the length becomes 17.0cm. Find (a) the spring constant and (b) its no-load length. (c) Calculate the energy stored in it when it is 17.0cm long. (d) Find the average force in the spring as it stretches from 10.0cm to 17.0cm. (e) Find the work done on the spring using this average force and the corresponding change in length. g=9.80m/s2.
3) The coil spring on one side of a car measures 36.0cm when a 75.0-kg lady sits on the fender exactly above that spring. When she places her 35.0-kg son on her lap, the spring gets further compressed to a length of 34.0cm. Find the spring constant and the no-load length of the spring, if the front mass of the car is 550kg. Suppose each spring carries half of the car's front weight. g =9.8m/s2.
4) A 90.0cm diameter wheel is spinning at 300 rpm in a vertical plane and has a handle sticking out on its edge (The wheel in the game "Price is Right" has several of such handles). Light shining straight down on it casts the shadow of this single handle on the ground that appears oscillation back and forth. Calculate (a) the radius and the angular speed (in rd/s) of the wheel, (b) its frequency, (c) its period, (d) the equation of motion of the shadow, if at t = 0, the shadow is at the rightmost point and the wheel is turning CCW, and (e) the position of the shadow relative to the midpoint at t = 0.0755s.
5) The equation of motion of certain waves arriving from ocean is y = [0.60m]cos(1.57t) judging by the up-and-down motion of a log on ocean's surface as measured by a physics student. For such waves, determine the (a) amplitude, (b) angular frequency, (c) frequency, and (d) period.
6) A student watching the in place oscillations of a log on a lake's surface notices that the log performs 20.0 full oscillations in 25.0seonds and he estimates the distance between the lowest and highest position of the log to be 50cm. For these oscillations, find the (a) period, (b) frequency, (c) angular frequency, (d) amplitude, and (e) the equation of motion of the log if he starts his stop watch when the log is at its highest position.
7) The graph of the back-and-forth motion of a piston in a car engine running at a constant rpm is shown in Fig. 13. Determine the (a) period, (b) frequency, (c) angular frequency, (d) amplitude, and (e) the engine's rpm. The total piston displacement in each full turn of the shaft is 14.0cm.
8) A 250.0-gram solid metal sphere attached to a spring of constant 256N/m is set into oscillation. Calculate (a) its angular frequency, (b) frequency, and (c) its period of oscillations.
The graph of the motion of a solid sphere hanging from a linear
spring of constant k = 2.15N/m is given in Fig. 14.
Determine the (a) period, (b) frequency, (c) angular frequency, (d) amplitude, and (e) mass of the sphere.
10) In Problem 9, find the magnitudes of maximum velocity and maximum acceleration and state where they occur.
Part 2: The Calculus-based Approach
The general form of the equation for oscillatory motion is x = A sin (ωt + φ) where φ is called the phase angle. Angle φ simply determines how far ahead or behind the oscillations are compared to x = A sin (ωt) taken as reference.
The equation of velocity can be obtained by taking the derivative of x = A sin (ωt + φ) with respect to time (t). Using v = dx/dt , the velocity v becomes:
v = Aω cos (ωt + φ)
Comparing this equation to v = vmax cos (ωt + φ), results in the value of maximum speed in SHM, simply vmax = Aω. It was discussed that vmax occurs at the midpoint (x = 0 ) as the oscillating object passes the equilibrium position.
To find the equation for acceleration, we need to use a = dv/dt that means taking the time derivative of the velocity equation. Doing this, results in
a = -Aω2 sin (ωt + φ)
Again, comparing this with a = amax sin (ωt + φ), results in the value of maximum acceleration in SHM, simply amax= -Aω2. It was discussed that amax occurs at the end points. When x = -A, the object is ready to go to the right and the acceleration is positive and maximum. When x = +A, the object is ready to go to the left, the acceleration has again the same highest magnitude, but toward the left, and therefore is negative and minimum.
Example 8: A 225gram small mass is hung from a linear spring and has stretched it by 6.4cm from its no-load length. Find (a) the spring constant. It is then further pulled down by 3.0cm and released. Find (b) its angular frequency, ω, (c) the frequency, f, (d) period, T , and the phase angle, φ of the resulting oscillations.
(a) In vertical direction, F = ky ; Mg = ky ; 0.225kg(9.81m/s2) = k (0.064m) ; k = 34.5 N/m.
(b) ω = SQRT( k/m) = [(34.5 N/m)/0.225kg]1/2 = 12.4 rd/s.
(c) ω = 2π f ; f = (12.4 rd/s) /2π ] = 1.97 Hz.
(d) T = 1/ f ; T = 1/ (1.97s-1) = 0.506s.
(e) To find φ, write the general form of the equation of motion for the oscillations: y = Asin(ωt + φ). We already know the value of y.
Substituting in the general form, we get : y = [0.030m] sin (12.4t + φ).
Since at t = 0 (time of release) y = -3.0cm = -0.030m ; therefore, by substituting these values into the last equation, we get:
-0.030m = [0.030]sin(12.4*0.0 + φ ) ; sin (12.4*0.0 + φ ) = -1 ; sin φ = -1 ; sin φ = sin (-90o).
This makes the acceptable answer to be φ = -90o. Plugging into the general form, the equation of the oscillations becomes:
y = 0.03 sin (12.4t - 1.57).
Note that the 1.57 means 1.57rd that is 3.14/2 or simply π/2 or 90o.
The Mass-Spring System (Calculus-Based Approach):
Referring to Fig. 15 for the horizontal oscillations of mass M by the linear spring k on a frictionless surface, we may think that as the mass is oscillating the force of the spring on it at any instant isFs = -kx.
This varying Fs generates a varying acceleration a in mass M that according to Newton's 2nd law is
Fs = Ma. Since a = d2x/dt2, we may write: Fs = Md2x/dt2 or,
-kx = Md2x/dt2 or, Md2x/dt2 + kx = 0 or, d2x/dt2 + (k/M) x = 0.
The solution to this differential equation gives x as a function of t :
x(t) = A sin (ωt + φ) where
Energy in Simple Harmonic Motion
As was discussed, for a mass spring system, when mass M is at its equilibrium position, it has its maximum speed, but the spring is neither compressed nor stretched. This means that, in this case, all energy(or the total energy of the system) is of kinetic type and potential energy is zero. On the other hand, when mass M is at an end point, since it momentarily comes to stop, its K.E. is zero, but its P.E. is maximum, because the spring is at its maximum stretch or compress. In other words, the same total energy is of potential type in this case. This can be shown mathematically as well. It is shown below:
Let U denote the potential (stored) energy in the spring: U = 0.5kx2. Since
x = A sin (ωt + φ), the expression for U becomes:
U = 0.5kA2sin2(ωt + φ). (1)
The kinetic energy, on the other hand is K.E. = 0.5Mv2.
Since v = Aω cos (ωt + φ), the expression for K.E. becomes:
K.E. = 0.5MA2ω2 cos2 (ωt + φ). (2)
Adding Equations (1) and (2) gives the total energy at any given instant. It is easy to verify that this total energy is equal to the maximum P.E. that is also equal to the maximum K.E..
Example 9: Add Equations (1) and (2) above to show that the total energy of an oscillator, at any given instant is
Total Energy = U + K.E. = 0.5kA2.
Solution: To be done by students.
Another good example of SHM is the "Simple Pendulum." Such device is made of a string connected to a tiny heavy and solid sphere hung from a fixed point and set into oscillation as shown (Fig. 16). It is easy to calculate the period of oscillation T of a simple pendulum as outlined below:
The length of a simple pendulum is 2.35m. Use g = 9.81 m/s2 to find the period of oscillation T of this pendulum for small angles.
Solution: Using T = 2π (L/g)1/2, we get: T = 6.28*(2.35/9.81)1/2 s. = 3.07s
Problem Set 2:
1) A mass attached to a spring oscillates according to the equation: x(t) = 24sin(16t + 0.50) where x is in cm and t in seconds. Find (a) the velocity and acceleration of the mass when x = 6.0cm for the earliest t > 0 and (b) the earliest time after t = 0.0 at which x = + 15cm and v < 0. click here
2) Mass M is attached to a linear spring of constant k on a horizontal frictionless surface. At t = 0.00, the mass is pulled to position x = +A and released. During the first cycle, at what positions and times do the following occur: (a) |v| = 0.500vmax, and (b) |a| = 0.500amax . Assume 3 sig. fig. on all variables. click here
3) The frequency of a mass-spring system set into oscillation is 2.50Hz. With an additional mass of 85.0 grams, the frequency reduces to 2.20Hz. Find mass M and the spring constant k. click here
4) The equation of oscillation of a mass-spring system is x(t) = 0.3cos(4t - 0.7) where x is in meters and t in seconds. The spring constant is 15N/m. Find (a) the amount of mass M, (b) the total energy, (c) the earliest time ( t >0) when K.E. = 0.5 P.E., and (d) the acceleration of the mass at t = 0.08s. Assume 3 sig. fig on all quantities. click here
5) The mass of an atom oscillating at a frequency of 8.6x1011Hz in its lattice is 1.2x10-26kg. If the amplitude of its oscillations is 0.045nm, find (a) its total energy in joules and eV, (b) its maximum speed, (c) its maximum acceleration, and (d) its equivalent spring constant. Note: 1eV = 1.6x10-19J.
6) What should be the length of a simple pendulum so that its period of oscillations is 2.00s? Let g = 9.81m/s2. click here
7) In an experiment to measure the acceleration of gravity ( g ) a simple pendulum of length 1.58m is used. The time for 100.0 oscillation is measured to be 254s. Calculate the "g" value in this experiment. click here
Radian, the SI unit of Angle:
One radian of angle is the central angle in any circle which opposite arc is equal to the radius of that circle. If a piece of string is cut equal to the radius R of a circle and then placed on the edge of that circle, as shown, the central angle corresponding (or opposite) to that arc is called one "radian."
As shown in Fig. 17, if a string
cut to length R,
is laid down on the edge of a circle as shown, it stretches from
A to B (Arc AB).
Example 11: Naming 3.14rd as " π ", calculate angles 360o, 180o, 90o, 60o, 45o, and 30o in terms of π.
|Fraction:||1 full circle||1/2 circle||1/4 circle||1/6 circle||1/8 circle||1/12 circle|
|Radians:||2π = 6.28rd||π||π/2||π/3||π/4||π/6|
The Arc Length-Central Angle Formula:
There is an easy formula that relates any central angle θ to its opposite arc s and the radius R of a circle. This formula is valid only if the central angle is measured or expressed in radians.
Example 12: Referring to Fig. 18, suppose angle θ is 148o and R = 1.25 in. Calculate the length of arc AB.
Solution: Using S = R θ, and converting degrees to radians, yields: S = (1.25 in.)(148o)( 3.14rd / 180o ) = 3.23 in.
The Angular Speed (ω):
The symbol "ω" is the lower case of the Greek letter Ω pronounced "omega." Angular speed ω is defined as the change in angle per unit of time. Mathematically, it may be written as
ω = dθ/dt. The preferred unit for angular speed is rd/s. The industrial unit for angular speed is rpm or revolutions per minute.
Note that each revolution or turn is 6.28 radians.
Example 13: A car tire spins at 240rpm. Calculate (a) its angular speed in rd /s, (b) the angle that any of its radii sweeps in 44 seconds, and (c) the arc-length that any point on its outer edge travels during this time knowing that R = 14 inches. Make sure that you completely solve this problem on paper using horizontal fraction bars everywhere.
Solution: 240rpm is the angular speed, ω. All we need to do in part (a) is to convert it from rpm to rd/s.
(a) ω = 240 (rev/min) = 240[6.28 rd/60s] = 25 rd/s. Note: 1rev = 6.28rd and 1min = 60s.
(b) ω = Δθ/Δt ; Δθ = ωΔt ; Δθ = [25 rd/s] (44s) ; Δθ = 1100 rd.
(c) s = Rθ ; s = 14 in.(1100 rd) = 15400 in. = 1283 ft = 0.24 miles.
The Linear Speed - Angular Speed Formula:
The same way that S and θ are related, we can develop a formula that relates v to ω. The formula is: v = Rω.
Writing these two similar formulas together helps their recall: s = Rθ ; v = Rω.
The derivation is easy. All you need to do is to first write s = Rθ as ds = Rdθ and then divide both sides of it by Δt.
We may reason that if s = Rθ, then Δs = RΔθ. Dividing through by dt, results in:
ds /dt = R dθ/dt (*)
As we see ds/dt is v or the linear speed. dθ/dt is ω or the angular speed; thus,
V = Rω.
Note: If you think Equation (*) is wrong because of division through by Δθ, refer to the end of the chapter for details.
Example 14: The radius of a car tire is 14 inches. Calculate (a) the linear speed v of any point on its outer edge if the tire spins at a constant angular speed of 25rd/s. (b) Find the linear distance or the arc length that any such point travels in 44s.
Solution: (a) Using v = Rω yields: V = (14 in.)(25 rd/sec) = 350 in/s.
(b) s = vt ; s = (350 in./s)(44s) = 15400in. Use horizontal fraction bars when you solve. Note: In Part (b), the equation x = (1/2)at2 + vi t is used in which a = 0 due to constant linear speed and x is replaced by s. We think of the arc length as a long string that is initially wrapped around the tire and as the tire rolls on the road, the string unwraps leaving the string on the road like a straight line (the distance traveled) for which the equation s = Vt or x = Vt is valid.
Test Yourself 2:
1) Radian is a unit of (a) length (b) angle (c) area. click here.
2) A central angle is an angle that has its vertex at the center of a (a) triangle (b) square (c) circle.
3) In any circle, the size of any central angle is equal to (a) the arc opposite to it when expressed in radians (b) half of the arc opposite to it (c) the radius of that circle.
4) Draw a circle and select two central angles in it, one equal to 90o and one equal to 45o. Since 90o is 1/4 of 360o, verify that the arc-length opposite to the 90o-angle you chose is also 1/4 of the whole circle. Also, verify that the 45o-angle you chose is opposite to an arc that is exactly 1/8 of the whole circle. What conclusion do you draw? Write it down. Is your conclusion in line with the correct answer to Question 3? click here.
5) 1rd is the central angle which opposite arc-length equals (a) the radius of the circle (b) the diameter of the circle (c) the perimeter of the circle.
6) The central angle which opposite arc equals the perimeter of the circle is (a) 360o (b) 2π radians (c) both a and b.
7) The central angle which opposite arc equals half of the circle is (a) 120o (b) 180o (c) 3.14radians (d) both b and c.
8) 2.00 radians of angle is equivalent to (a) 114.6o (b) 120o (c) 170o. click here.
9) S = Rθ calculates the arc length that is opposite to central angle θ only if θ is expressed in (a) degrees (b) radians (c) grads.
10) The arc length opposite to a 2.00rd central angle in a circle of radius R = 6.00" is (a) 12.0" (b) 8.00" (c) 18.0".
11) The arc length opposite to a 114.6o central angle in a circle of radius R = 6.00" is (a) 18.0" (b) 28.0" (c) 12.0".
12) Angular speed is defined as (a) the arc length traveled per unit of time (b) the angle traveled per unit of time (c) the angle swept by by any one radius per unit of time (d) both b and c. click here.
13) If angular change is Δθ and time change is Δt, then angular speed, ω, is (a) Δθ /Δt (b) Δt /Δθ (c) ΔθΔt.
14) The angular speed formula ω = Δθ /Δt is the counterpart of the linear speed formula v = Δx /Δt. (True) or (False)?
15) A spoke on a bicycle wheel travels or sweeps 150rd of angle every minute. Its angular speed is (a) 150rd/s (b) 150rd/min (c) 2.50rd/s (d) both b and c. click here.
16) RPM is (a) a unit of angle (b) a unit of angular acceleration (c) an industrial unit for angular speed.
17) rd/s is (a) an industrial unit for angle (b) the SI unit for angular speed (c) neither a nor b. click here.
18) 180 rpm is the same thing as (a) 3.0 rps (b) 3.0 revolutions per second (c) 3.0 turns per second (d) a, b, and c.
19) 240 rpm is the same thing as (a) 240 turns per minute (b) 4.0 turns per second (c) both a and b.
20) 1 rpm is the same thing as 6.28 rd/min. True or False? click here.
21) 3600 rpm is equivalent to (a) 60 rev/s (b) 60x6.28rd /s (c) 377 rd/s (d) a, b, and c.
22) A helicopter propeller rotates at 956 rpm. Its angular speed is (a) 100 rd/s (b) 200 rd/s (c) 300 rd/s. click here.
23) The formula that relates linear speed to angular speed is (a) s = Rθ (b) v = Rω (c) x = vt + 1/2 at2.
24) A helicopter propeller rotates at 956 rpm. The linear speed of the tip of its propeller that is 5.00m long is (a) 500m/s (b) 750m/s (c) 956m/s. click here.
25) A helicopter propeller rotates at 956 rpm. The linear speed of the midpoint of its propeller that is 2.50m from its axis of rotation is (a) 500m/s (b) 750m/s (c) 250m/s.
26) A helicopter propeller rotates at 956 rpm. The angular speed of the midpoint of it is (a) the same as the angular speed of the tip of it (b) is 1/2 of the angular speed of the tip of it (c) neither a nor b. click here.
27) All points on a rotating solid wheel have the same (a) linear speed (b) acceleration (c) angular speed.
28) The linear speed of a point on a solid rotating disk (a) depends on R, its distance from the center of rotation (b) does not depend on R (c) depends on the angular speed of the disk. (d) both a and c.
29) All points on the outer edge of a rotating solid wheel have (a) the same angular speed (b) the same linear speed (c) the same linear velocity (d) a and b. click here.
30) A solid wheel of radius 0.400m spins at 478 rpm. Find its angular speed in rd/s as well as the linear speed of points on it that are at radii 0.100m, 0.200m, and 0.300m.