Chapter 10
Simple Harmonic Motion (SHM):
A mass attached to a linear spring and set into upanddown motion performs a motion that is called " simple harmonic motion, or SHM. " We need to first study the behavior of a linear spring.
Linear Springs:
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 , the proportionality constant, is called the " spring constant."
Example 1: A linear spring has an unstretched length of 18.0cm. When it is under a load of 125N, its total length is 20.5cm. Calculate (a) its constant, k, and (b) the load that makes it 25.0cm long.
The Metric unit for k is N/m.
Example 2: A linear spring has a length of 35.0cm when under a load of 225N and a length of 43.0cm when under a load of 545N. Find (a) its constant, and (b) its free (no load) length.
Solution: To be solved by students.
The Linear Spring Formula:
Note that the formula ΔF = k Δx is a relation between the applied force (to the spring) and the change in the spring's length. The spring force ( F_{s} ) is always opposite to the applied force. As the following figures indicate,
when F_{applied} is to the right, Δx is positive, F_{s} pulls to the left and is negative, and
when F_{applied} is to the left, Δx is negative , F_{s} pushes to the right and is positive.
Based on the above F_{s} =  k x is the formula for a linear spring. When ( x ) is positive, F_{s} is negative and vise versa. This fact is reflected by the (  ) sign in the formula. Note that F_{s} is not the applied force, it is the force that the spring exerts.
Simple Harmonic Motion:
Recall the definition of angular speed ω: ω = θ/t ; therefore, θ = ωt. This will be used later.
If mass M performs a uniform circular motion in a vertical plane, its shadow on the xaxis performs a backandforth motion that is called simple harmonic motion. To understand the following figure, visualize that mass M moves slowly and counterclockwise along the circle (of radius A), and at different positions, picture its shadow on the floor. The angular position of mass M on the circle is determined by θ. Corresponding to every θ there is a shadow position measured by x from C to H (or O to K). It is possible to relate x to θ. Since θ = ωt ; therefore, x can be related to ωt.
The graph of x vs. θ is shown below with the assumption that at t = 0, θ = 0. That is mass M starts its motion from the position for which θ = 0. Note that the farthest Point K can go from Point O is as much as length A, the radius of the circle. A, the maximum deviation from the equilibrium position, is called the " Amplitude " of oscillations.
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 = 1.77 seconds.
Solution: (a) ω = 180 (rev / min) ( 6.28rd / rev)( min / 60 sec ) = 18.8 rd /s
(b) The constants are known: ω = 18.8 rd /s and A = 30.0 cm ; The variables are x and t .
The equation of motion for oscillations is : x = A cos(ωt) ; x = [30.0cm] cos (18.8t ) ; ( t ) in seconds.
(c) Substituting for t = 1.77sec, x becomes: x = (30.0cm) cos(18.8*1.77 rd ) = 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 = A cos(ωt), it is clear that
(a) A = 3.23 cm ; (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.
The MassSpring System:
As it was mentioned, when mass M attached to a linear spring is pulled and released, its upanddown 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. Figure (a) below shows a spring that is not loaded. Figure (b) shows the same spring but loaded and stretched a distance (  h ), and Figure (c) shows the loaded spring stretched further a distance ( A ) and released. It shows that the attached mass M oscillates up and down to (+A) and (A) above and below the equilibrium level.
Example 5: A 102gram mass hung from a weak spring has stretched it by 3.00cm. Let g = 9.81m/s^{2} and calculate (a) the load on the spring and (b) the spring constant in N/m. If the massspring system is initially in static equilibrium and motionless, 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/s^{2}) = 1.00N.
(b) ΔF = k Δx ; k = (1.00N) /( 0.0300m) = 33.3 N/m.
(c) ω = SQRT( k / M ) = SQRT [( 33.3 N/m ) / (0.102 kg)] = 18.1 rd/s.
(d) f = ω / (2π ) ; f = 2.88 Hz.
(e) T = 1 / f ; T = 0.347s.
( f ) The 2.00cm that the mass is pushed up above its equilibrium level, initially, becomes its amplitude. A = +2.00cm.
(g) Knowing the constants A = 2.00cm and 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 released from X = +2.00cm.
Example 6: The graph of x ( the distance from the equilibrium position ) versus time ( t ) for the oscillations of a massspring system is given below:
For such oscillations, find (a) the amplitude, (b) the period, (c) the frequency, (d) the angular speed (frequency), (e) the spring constant ( k ) if the mass of the object is 250 grams, and ( f ) the equation of motion for the oscillations.
Solution: (a) A = 2.00cm ; (b) T = 2 (0.125s) = 0.250 s ; (c) f = 1 / T ; f = 4.00 Hz.
(d) ω = 2π f ; ω = 2π (4.00/s) = 25.1 rd/s.
(e) ω = ( k / M)^{(1/2)} ; ω^{2} = ( k / M) ; k = Mω^{2} ; k = (0.250kg)(25.1 rd/s)^{2}_{ } ;_{ }k = 158 N/m.
(f) x = A sin (ωt) ; x = [2.00cm] sin ( 25.12t ). The given graph is a sine function.
Note that at t = 0, X = 0, according to the given graph. It is a sine function that is zero at t =0. and not a cosine function.
Linear Velocity and Acceleration in Simple Harmonic Motion:
Velocity: If an object is oscillating back and forth on a frictionless surface as shown below, it is easy to see that its velocity becomes zero at the extreme ends. This is simply because it has to come to stop at those points to return. It is also easy to see that velocity gains its maximum magnitude at the midpoint or the equilibrium level. We may therefore state that:
"In SHM, maximum speed occurs at x = 0 (the equilibrium level), and zero speed occurs at each of the extreme ends (x = +A or A)."
Acceleration: As shown above, at the middle (x = 0), a = 0 because the spring is neither stretched or compressed, F = 0. At the extreme ends, when the spring is at its maximum stretch or compress, the spring force is at its maximum magnitude, and therefore the acceleration it gives to the attached mass is maximum in magnitude as well. We may therefore state that:
"In SHM, a_{max.} occurs at x = +A or A, (at the extreme ends where the force magnitude is maximum), and zero acceleration occurs at x = 0, the midpoint."
Using Calculus, the equations for x, v, and a in SHM derive as follows:
x_{max.}, v_{max.} , and a_{max.} can be determined by setting each of cos(ωt) and sin(ωt) above equal to its maximum value that is 1. x_{max.}, v_{max.} , and a_{max.} become:
x_{max} = A , v_{max.} =  Aω and a_{max.}_{ }= Aω^{2}.
We may ignore the (  ) signs if only the magnitudes are concerned, as shown above. Note that x_{max} = A = amplitude.
Example 7: The equation of motion of a 22kg 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), frequency, period, maximum speed, maximum acceleration (magnitude), and its position at to t = 0.19s.
Solution: A = 1.2m ; ω = 3.14 rd/s ; f = ω / (2π) = 0.50 s^{1} ; T = 1 / f = 2.0s.
V_{max}  = Aω ; V_{max}  = (1.2m)(3.14 rd/s) = 3.8 m/s (occurs at the middle).
a_{max}  = Aω^{2 } ; a_{max}  _{ }= (1.2m)(3.14rd/s)^{2} = 11.8 m/s^{2}.
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.
Chapter 10 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) F^{2} = k(Δx) (c) F = k(Δx).
2) The formula for the applied force, F_{appl.} to a spring and the change in its length Δx is (a) F_{appl.} = k(Δx) (b) F_{appl.} =  k(Δx) (c) neither a, nor b.
3) The relation between the spring force, F_{s} to Δx is (a) F_{s} = k(Δx) (b)  F_{s} =  k(Δx) (c) F_{s} =  k(Δx).
4) The spring formula F_{s} =  k(Δx) is normally written as F_{s} =  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 21.58kg load has a constant of 20.N/cm (b) 15N/m (c) 30.N/cm.
For the following questions, refer to the figure under "Simple Harmonic Motion" 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) 0.00cm (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) The equation of the shadow's oscillations is (a) x = [10.0cm] * cos ( 0.523 t ) (b) x = (1/2)(0.523 t^{2}) (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? In cos(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 = 2π f , the value f, (number of turns per second), is (a) f = 0.0833 s^{1} (b) f = 0.0833 /s (c) f = 0.0833 Hz (d) f = 0.0833 Hz (e) a, b, c, & d.
21) Based on 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 info 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(18.78 t) where x is in cm and t is in seconds. Based on the equation, click here .
23) the amplitude of oscillations is (a) 8.00cm (b) 4.00cm (c) 2.00cm.
24) the angular frequency ω that is the # of radians traveled (or repeated) per second is (a) 18.78 rd/min (b) 18.78 rd/s (c) 18.78 degrees per second. click here .
25) The frequency of oscillations, f (the # of turns per second or the # of backandforth motions 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 position of the object from its equilibrium position at t = 0 is (a) 2.00cm (b) 6.00cm (c) 4.00cm.
28) The position of the object from its equilibrium position at t = 0.0175s is (a) 3.786cm (b) 3.21cm (c) 4.00cm.
29) The position of the object from its equilibrium position at t = 2.945s is (a) 2.28cm. (b) 1.293cm. (c) 1.82cm.
Problem: Refer to the figure of Example 6. There are 12 segments (time intervals) on the taxis. 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 you calculator in the correct mode. click here
30) The time interval corresponding to each segment is (a) .04167s (b) 0.02083s (c) 0.01042s.
31) Each vertical line under the sinecurve shows the distance of the object from (a) the equilibrium position. (b) the midpoint. (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 the equilibrium at t = 1(0.02083s) is (a) 2.00cm. (b) 1.00cm. (c) 3.00cm.
34) The distance from the equilibrium at t = 2(0.02083s) is (a) 1.20cm. (b) 1.73cm. (c) 0. click here
35) The distance from the equilibrium at t = 3(0.02083s) is (a) 0.80cm. (b) 2.00cm. (c) 0.
36) The distance from the equilibrium at t = 4(0.02083s) is (a) 0.60cm. (b) 1.73cm. (c) 0.50.
37) The distance from the equilibrium 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 = Aωt^{2}.
39) A, the amplitude, in oscillatory motion is the same thing as R, the radius, in circular motion. (a) True (b) False
40) In Example 6, 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 6, the  a_{max.} of the mass is (a) Aω^{2} (b) Aα t (c) Aα t^{2}.
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. For an oscillating massspring system, for example, the reason is that (a) the velocity at end points may not become 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:
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 200.0gram mass hanging on a linear spring gives it a length of 14.0cm. When the hanging load is 350.0 grams, the length becomes 17.0cm. Find (a) the spring constant and (b) its noload 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 by this average force through the corresponding change in length. g =9.8m/s^{2}.
3) The coil spring on one side of a car measures 36.0cm when a 75.0kg lady sits on the fender exactly above that spring. When she places her 35.0kg son on her lap, the spring gets further compressed to a length of 34.0cm. Find the spring constant and the noload length of the spring, if the front mass of the car is 550.0kg. Suppose each spring carries half of the car's front weight. g =9.8m/s^{2}.
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) its equation of motion, if at t=0, that shadow is at the rightmost point and the wheel is turning CCW, and (e) the position of the shadow at t=0.0755s.
5) The equation of motion of certain waves arriving from the ocean is y = [0.60m]cos(1.57t) judging by the upanddown 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) the 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 50.cm. For these oscillations, find the (a) period, (b) frequency, (c) angular frequency, (d) amplitude, and (e) the equation of motion. He starts his stop watch when the log is at maximum height.
7) The graph of the motion of a piston in a car engine at a constant rpm is shown on the right. Determine the (a) period, (b) frequency, (c) angular frequency, (d) amplitude, if the total piston displacement is 14.0cm, and (e) the rpm the engine is at.


8) A 250.0gram 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.
9) The
graph of the motion of a solid sphere hanging from a linear spring of
constant k = 12.15N/m is given by the figure on the right. 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.
Answers:
1) 20.0cm, 40.0cm 2) 49N/m, 10. cm, 0.12J, 1.715N, 0.12J
3) 17150N/m, 56.0cm
4) 45.0cm & 31.4 rd/s, 5.00Hz, 0.200s, x = [0.45m] cos( 31.4t ), 32.3cm
5) 0.60m, 1.57 rd/s, 0.25Hz, 4.0s
6) 1.25s, 0.800s^{1}, 5.02rd/s, 25.0cm, y = [0.25m]cos(5.02t)
7) 0.0250s, 40.0 Hz, 251 rd/s, 7.00cm, 2400rpm. 8) 32.0 rd/s, 5.10Hz, 0.196s
9) 0.900s, 1.11 Hz, 6.97rd/s, 3.00cm, 250.grams,
10) V_{max} = 0.209m/s at y = 0, and a_{max} = 1.46m/s^{2} at y = (+/) 3.00cm