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PHYS1100 Practice problem set, Chapter 14: 4, 8, 13, 16, 18, 22, 35, 36, 43, 61, 67, 75

14.4. Model: The air-track glider attached to a spring is in simple harmonic motion.

Visualize: The position of the glider can be represented as x(t) = A cos ωt.

Solve: The glider is pulled to the right and released from rest at 0 st = . It then oscillates with a period 2.0 sT =

and a maximum speed max40 cm s 0.40 m sv = = .

(a) max

max

2 2 0.40 m s and rad s 0.127 m 12.7 cm

2.0 s rad s

vv A A

T

π πω ω π

ω π= = = = ⇒ = = = =

(b) The glider’s position at t = 0.25 s is

( ) ( )( )0.25 s0.127 m cos rad s 0.25 s 0.090 m 9.0 cmx π = = =

14.8. Visualize: A phase constant of 23π− implies that the object that undergoes simple harmonic motion is in

the third quadrant of the circular motion diagram. That is, the object is moving to the right.

Solve: The position of the object is given by the equation

( ) ( ) ( ) ( ) ( ) 20 0 3

cos cos 2 6.0 cm cos rad s radx t A t A ft tω φ π φ π π = + = + = −

The amplitude is A = 6 cm and the period is 1 2.0 s.T f= = With 0

2 3 rad,φ π= − x starts at 12

A− and is moving

to the right (getting more positive).

Assess: As we see from the graph, the object starts out moving to the right.

14.13. Model: The air-track glider attached to a spring is in simple harmonic motion.

Solve: Experimentally, the period is ( ) ( )12.0 s 10 oscillations 1.20 s.T = = Using the formula for the period,

( )2 2

2 22 0.200 kg 5.48 N m

1.20 s

mT k m

k T

π ππ

= ⇒ = = =

2

14.16. Model: The mass attached to the spring is in simple harmonic motion.

Solve: (a) The period is

( )( )

0.507 kg2 2 1.00 s

20 N m

mT

kπ π= = =

(b) The angular frequency is 2 2 1.0 s 2 rad s.Tω π π π= = =

(c) To calculate the phase constant 0

φ ,

( ) ( )1 1 10 0 0 0 2 3

cos 0.05 m 0.10 m cos cos radx A φ φ φ π−= ⇒ = ⇒ = = ±

Because the mass is moving to the right at t = 0 s, it is in the lower half of the circular motion diagram. Hence, 1

0 3 rad.φ π= −

(d) At t = 0 s, ( ) ( )( ) ( )00.10 m 2 rad s sin 2

xv t tπ π φ= − + becomes

( )( ) ( )13

0.10 m 2 rad s sin rad 0.544 m sπ π− − =

(e) The maximum speed ( )( )max2 rad s 0.10 m 0.628 m sv Aω π= = = .

(g) At t = 1.3 s, ( ) ( ) 11.3 s 3

0.10 m cos 2 1.3 s rad 0.0669 mx π π = − = .

(h) At t = 1.3 s, ( ) ( )( ) ( )( ) 1

31.3 s0.10 m 2 rad s sin 2 rad s 1.3 s rad 0.467 m s

xv π π π = − − = − .

14.18. Model: The vertical oscillations constitute simple harmonic motion.

Visualize:

Solve: (a) At equilibrium, Newton’s first law applied to the physics book is

( )

( )( ) ( )

sp

2

0 N 0 N

0.500 kg 9.8 m s 0.20 m 24.5 N m

yF mg k y mg

k mg y

− = ⇒ − ∆ − =

⇒ = − ∆ = − − =

(b) To calculate the period:

24.5 N m 2 2 rad7.0 rad s and 0.898 s

0.50 kg 7.0 rad s

kT

m

π πω

ω= = = = = =

(c) The maximum speed is

( )( )max0.10 m 7.0 rad s 0.70 m sv Aω= = =

Maximum speed occurs as the book passes through the equilibrium position.

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14.22. Model: Assume a small angle of oscillation so there is simple harmonic motion.

Solve: The period of the pendulum is

00

2 4.0 sL

Tg

π= =

(a) The period is independent of the mass and depends only on the length. Thus 04.0 s.T T= =

(b) For a new length L = 2L0,

00

22 2 5.66 s

LT T

gπ= = =

(c) For a new length L = L0/2,

00

2 12 2.83 s

2

LT T

gπ= = =

(d) The period is independent of the amplitude as long as there is simple harmonic motion. Thus 4.0 s.T =

14.35. Model: The vertical mass/spring systems are in simple harmonic motion.

Visualize: Please refer to Figure P14.35.

Solve: (a) For system A, the maximum speed while traveling in the upward direction corresponds to the

maximum positive slope, which is at t = 3.0 s. The frequency of oscillation is 0.25 Hz.

(b) For system B, all the energy is potential energy when the position is at maximum amplitude, which for the first time

is at t = 1.5 s. The time period of system B is thus 6.0 s.

(c) Spring/mass A undergoes three oscillations in 12 s, giving it a period A

4.0 s.T = Spring/mass B undergoes 2

oscillations in 12 s, giving it a period B

6.0 s.T = We have

A B A A BA B

A B B B A

4.0 s 22 and 2

6.0 s 3

m m T m kT T

k k T m kπ π

= = ⇒ = = =

If mA = mB, then

B A

A B

4 92.25

9 4

k k

k k= ⇒ = =

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14.36. Model: The astronaut attached to the spring is in simple harmonic motion.

Visualize: Please refer to Figure P14.36.

Solve: (a) From the graph, T = 3.0 s, so we have

( )2 2

3.0 s2 240 N m 54.7 kg

2 2

m TT m k

kπ

π π

= ⇒ = = =

(b) Oscillations occur about an equilibrium position of 1.0 m. From the graph, ( )102

0.80 m 0.40 m, A φ= = =

0 rad, and

2 22.094 rad s

3.0 sT

π πω = = =

The equation for the position of the astronaut is

( ) ( ) ( )

( ) ( ) ( )

cos 1.0 m 0.4 m cos 2.094 rad s 1.0 m

1.2 m= 0.4 m cos 2.094 rad s 1.0 m cos 2.094 rad s 0.5 0.50 s

x t A t t

t t t

ω = + = +

⇒ + ⇒ = ⇒ =

The equation for the velocity of the astronaut is

( ) ( )

( )( ) ( )( )0.5 s

sin

0.4 m 2.094 rad s sin 2.094 rad s 0.50 s 0.725 m s

xv t A t

v

ω ω= −

⇒ = − = −

Thus her speed is 0.725 m/s.

14.43. Model: The ball attached to a spring is in simple harmonic motion.

Solve: (a) Let t = 0 s be the instant when x0 = −5 cm and v0 = 20 cm/s. The oscillation frequency is

2.5 N m5.0 rad / s

0.10 kg

k

mω = = =

Using Equation 14.27, the amplitude of the oscillation is

( )22

22 00

20 cm /s5 cm 6.40 cm

5 rad / s

vA x

ω

= + = − + =

(b) The maximum acceleration is 2 2

max160 cm sa Aω= = .

(c) For an oscillator, the acceleration is most positive ( )maxa a= when the displacement is most negative

( )maxx x A= − = − . So the acceleration is maximum when 6.40 cmx = − .

(d) We can use the conservation of energy between 0

5 cmx = − and x1 = 3 cm:

( )2 2 2 2 2 2 21 1 1 10 0 1 1 1 0 0 12 2 2 2

0.283 m s 28.3 cm sk

mv kx mv kx v v x xm

+ = + ⇒ = + − = =

Because k is known in SI units of N/m, the energy calculation must be done using SI units of m, m/s, and kg.

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14.61. Model: A completely inelastic collision between the two gliders resulting in simple harmonic motion.

Visualize:

Let us denote the 250 g and 500 g masses as m1 and m2, which have initial velocities vi1 and vi2. After m1 collides

with and sticks to m2, the two masses move together with velocity vf.

Solve: The momentum conservation equation f ip p= for the completely inelastic collision is ( )1 2 f

m m v+ =

1 i1 2 i2m v m v+ . Substituting the given values,

( ) ( )( ) ( )( )f f0.750 kg 0.25 kg 1.20 m s 0.50 kg 0 m s 0.40 m sv v= + ⇒ =

We now use the conservation of mechanical energy equation:

( ) ( ) ( )

( )

2 21 1s s 1 2 f2 2compressed equilibrium

1 2f

0 J 0 J

0.750 kg0.40 m s 0.110 m

10 N m

K U K U kA m m v

m mA v

k

+ = + ⇒ + = + +

+⇒ = = =

The period is

1 2 0.750 kg2 2 1.72 s

10 N m

m mT

kπ π

+= = =

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14.67. Model: Assume that marble hoop

r R<< and that θ is a small angle.

Visualize:

Solve: The marble is like an object on an inclined plane. The net force on the marble in the tangential direction is

2 2

2 2sin sin

d dw ma mR mR mg mR

dt dt

θ θθ α θ− = = = ⇒ − =

where α is the angular acceleration. With the small-angle approximation sinθ θ≈ , this becomes

22

2

d g

Rdt

θθ ω θ= − = −

This is the equation of motion of an object in simple harmonic motion with a period of

22

RT

g

ππ

ω= =

14.75. Model: The two springs obey Hooke’s law.

Visualize:

Solve: There are two restoring forces on the block. If the block’s displacement x is positive, both restoring forces

– one pushing, the other pulling – are directed to the left and have negative values:

( ) ( ) ( ) ( )net sp 1 sp 2 1 2 1 2 effx x xF F F k x k x k k x k x= + = − − = − + = −

where eff 1 2k k k= + is the effective spring constant. This means the oscillatory motion of the block under the

influence of the two springs will be the same as if the block was attached to a single spring with spring constant keff.

The frequency of the blocks, therefore, is

2 2eff 1 2 1 21 22 2

1 1

2 2 4 4

k k k k kf f f

m m m mπ π π π

+= = = + = +