Chapter 15 Answers

Test Yourself 1

 1) c

 2) a

 3) c

 4) a

 5) c

 6) c

 7) b

 8) c

 9) a

 10) a

 11) c

 12) a

 13) b

 14) c

 15) c

 16) a

 17) b

 18) c

 19) a

 20) e

 21) a

 22) a

 23) b

 24) b

 25) d

 26) a

 27) c

 28) a

 29) b

 30) b

 31) c

 32) c

 33) b

 34) b

 35) b

 36) b

 37) b

 38) a

 39) a

 40) c

 41) c

 42) a

 43) d

 44) d

 

Answers To Problem Set 1

1) 20.0cm, 40.0cm    2) 49N/m,  10cm,  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,  44.2grams

10) Vmax = 0.209m/s at y = 0,   and   amax = 1.46m/s2 at y = (+/-) 3.00cm

Answers to Problem Set 2

1) (a) At t = 0.15s,  x = 6.0cm,  v = -370 cm/s, and  a = -1500cm/s2.    (b) t = 0.12s

2)  (a) T = 6.28/ω.  t1 = 0.524/ω,  t2 = 2.616/ω, t3 = 3.66/ω, and  t4 = 5.76/ω.

                                            x1 = 0.866A,  x2 = -0.866A, x3 = -0.866A, and  x4 = 0.866A.

     (b)  t1 = 1.05/ω,  t2 = 2.09/ω, t3 = 4.19/ω, and  t4 = 5.23/ω.

            x1 = 0.500A,  x2 = -0.500A, x3 = -0.500A, and  x4 = 0.500A

3)   M = 0.292 kg,  k = 72.3N/m.

4)  (a) 0.938kg,  (b) 0.675J,  (c) 0.0213s,  (d) -4.46m/s2.

5)  (a) 3.54x10-22J, 0.0022eV  (b) 243m/s,  (c) 1.3x1015m/s2,  (d) 0.35N/m.

6)   99.5cm.    7)   9.67m/s2

Test Yourself 2

 1) b

 2) c

 3) a

 4) In any circle, any central angle equals its opposite arc, angle-wise.

 5) a

 6) c

 7) d

 8) a

 9) b

 10) a

 11) c

 12) d

 13) a

 14) True

 15) d

 16) c

 17) b

 18) d

 19) c

 20) True

 21) d

 22) a

 23) b

 24) a

 25) c

 26) a

 27) c

 28) d

 29) d

 30) 50.0rd/s, 5.00m/s, 10.0m/s, & 15.0m/s

Solution to Set 2 Problems:

1. (a):  x(t) = 24sin(16t + 0.50)   ;    6 = 24sin(16t + 0.50)

            sin(16t + 0.50) = 0.25    ;     sin-1 (0.25) = 0.251  ; therefore,

            sin(16t + 0.50) = sin (0.251 rd)

Either   16t + 0.50 = 2nπ + 0.251  in which n = -1 and n = 0 do not produce a positive value for t , but n = 1 gives us the first possible positive time t = 0.377s.

Or,        16t + 0.50 = 2nπ + π - 0.251 in which n = -1 does not produce a positive value for t , but n = 0 gives us the first possible time t = 0.149s.

The velocity and acceleration at this earliest time become:

v(t) = dx/dt = 16x24cos(16t + 0.50)    or,    v(0.149) = - 370cm/s.

a(t) = dv/dt = -162x24sin(16t + 0.50)    or,    a(0.149) = - 1500cm/s2.

1 (b): x(t) = 24sin(16t + 0.50)   ;    15 = 24sin(16t + 0.50)

                    sin(16t + 0.50) = 0.625    ;     sin-1 (0.625) = 0.62513  ; therefore,

            sin(16t + 0.50) = sin (0.62513 rd)

Either   16t + 0.50 = 2nπ + 0.62513  in which n = -1 does not produce a positive value for t , but n = 0 gives us the first possible positive time t = 0.322s.

Or,        16t + 0.50 = 2nπ + π - 0.62513 in which n = -1 does not produce a positive value for t , but n = 0 gives us the first possible time t = 0.126s.

The velocity equation is v(t) = dx/dt = 16x24cos(16t + 0.50).

v(0.322) = 310 m/s > 0 and not acceptable.

v(0.126) = -311 m/s < 0 and acceptable; therefore the answer is t = 0.126s.

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2 (a):  Since at t = 0, the object is at +A ; therefore, 

x(t) = Acos(ωt),  and consequently   v = -Aω sin(ωt).     We want:

|v| = 0.5vmax.    This absolute value equation is in fact 2 equations:

Either  v = +0.5vmax   or    v = - 0.5vmax.

If  v = +0.5vmax  that means  -Aω sin(ωt) = +0.5Aω    or   sin(ωt) = - 0.5   or

sin(ωt) = sin(-0.524).   Its 2 solutions are:  ωt = 2nπ -0.524  &  ωt = 2nπ + π +0.524

Giving integer n values like -1, 0, +1, and ...., possible answers from these 2 solutions are:

 t = 5.76/ω  and  t = 3.66/ω.  (2 solutions)

If  v = - 0.5vmax   that means  -Aω sin(ωt) = -0.5Aω    or   sin(ωt) = 0.5   or

sin(ωt) = sin(0.524).    Its 2 solutions are: ωt = 2nπ +0.524  &  ωt = 2nπ + π -0.524

Possible answers are : t = 0.524/ω  and  t = 2.62/ω.  (Another 2 solutions)

Positions: To find positions, just substitute the following values for ωt in the position equation { x = Acos(ωt) } as follows: ωt = 2.09,  4.19,  1.05,   and   5.23

For ωt = 0.524we get  x(t) = Acos(0.524) =   0.866A.

For ωt = 2.62,      we get  x(t) = Acos(2.09) = -0.866A.

For ωt = 3.66,      we get  x(t) = Acos(3.66) = -0.866A.

For ωt = 5.76,      we get  x(t) = Acos(5.76) =   0.866A.

 

2 (b): From  x(t) = Acos(ωt) v = -Aω sin(ωta = -Aω2 cos(ωt). 

We want: |a| = 0.5amax.    This absolute value equation is in fact 2 equations:

Either  a = +0.5amax   or    a = - 0.5amax.

If  a = +0.5amax  that means  -Aω2 cos(ωt) = +0.5Aω2    or   cos(ωt) = - 0.5   or

cos(ωt) = cos(2.09).   Its 2 solutions are:  ωt = 2nπ +2.09  &  ωt = 2nπ -2.09.

Giving integer n values like -1, 0, +1, and ...., possible answers from these 2 solutions are:

 t = 2.09/ω  and  t = 4.19/ω.  (2 solutions)

 If  a = -0.5amax  that means  -Aω2 cos(ωt) = -0.5Aω2    or   cos(ωt) =  0.5   or

cos(ωt) = cos(1.05).   Its 2 solutions are:  ωt = 2nπ +1.05  &  ωt = 2nπ -1.05.

Giving integer n values like -1, 0, +1, and ...., possible answers from these 2 solutions are:

 t = 1.05/ω  and  t = 5.23/ω.  ( Another 2 solutions)

Positions: To find positions, just substitute the following values for ωt in the position equation { x = Acos(ωt) } as follows: ωt = 2.09,  4.19,  1.05,   and   5.23

For ωt = 1.05, we get  x(t) = Acos(1.05) =   0.5A.

For ωt = 2.09, we get  x(t) = Acos(2.09) = -0.5A.

For ωt = 4.19, we get  x(t) = Acos(4.19) = -0.5A.

For ωt = 5.23, we get  x(t) = Acos(5.23) =   0.5A.

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4. (a)   ω = (k/M)1/2  or  M = ω2/k  or  M = [42/15]kg = 0.938kg.

(b)       U = 0.5kA2 = 0.5(15)(.3)2 J = 0.675J

(c)  Since x(t) = 0.3cos(4t - 0.7),  velocity becomes:  v(t) = -1.2sin(4t - 0.7).

K.E. = 0.5 P.E. means:  (1/2)Mv2 = (1/2) (1/2)kx2  or, 

(1/2)M[-1.2sin(4t - 0.7)]2 =  (1/2) (1/2)k[0.3cos(4t - 0.7)]2.   Simplifying, we get:

tan2(4t - 0.7) = 0.5.    (Note: ω = (k/M)1/2  or,  k = Mω2  & ω = 4 from the equation.)

or,     tan(4t - 0.7) = ± 0.707  (2 equations)

For  tan(4t - 0.7) = + 0.707  we write:  tan(4t - 0.7) =  tan(0.615)  that gives us:

4t - 0.7 = nπ + 0.615.   n = -1 produces negative t.  

But, n = 0 results in one possible solution!  t = 0.329s.

For  tan(4t - 0.7) = - 0.707  we write:  tan(4t - 0.7) =  tan(-0.615)  that gives us:

4t - 0.7 = nπ - 0.615.   n = -1 produces negative t.  

But, n = 0 results in the desired solution:  t = 0.0213s. (The earliest)

4 (d) x(t) = 0.3cos(4t - 0.7)  v(t) = -1.2sin(4t - 0.7) a(t) = -4.8cos(4t - 0.7)

a(0.08) = -4.8cos(4*0.080 - 0.7) = -4.46 m/s2.