Question 1

Exhibit 15-1
A graph of position versus time for an object oscillating at the free end of a horizontal spring is shown below.A point or points at which the object has positive velocity and zero acceleration is(are)  Use this exhibit to answer the following question(s).
-Refer to Exhibit 15-1.The point at which the object has negative velocity and zero acceleration is
A)A
B)B
C)C
D)D
E)E

Accepted Answer

According to the graph, at point C, the object has a negative velocity (slope is negative) and zero acceleration (slope is zero).

Question 2

A 2.00 m-long 6.00 kg ladder pivoted at the top hangs down from a platform at the circus.A 42.0 kg trapeze artist climbs to a point where her center of mass is at the center of the ladder and swings at the system's natural frequency.The angular frequency (in s−¹)of the system of ladder and woman is A)1.01. B)3.07. C)4.03. D)8.05. E)16.2.

Accepted Answer

The natural frequency of a simple pendulum is given by:
ω = √(g/l)
where g is the acceleration due to gravity and l is the length of the pendulum.
In this case, the ladder and trapeze artist form a simple pendulum with a length of 2 meters. The trapeze artist is added at the midpoint of the ladder, so the effective length of the pendulum is 1 meter. 
The total mass of the system is 6+42 = 48 kg. 
The center of mass of the ladder is at a distance of 1 meter from the top, so the torque due to the weight of the ladder is:
τ_l = (6 kg)·(9.81 m/s2)·(1 m) = 58.86 N·m
The torque due to the weight of the trapeze artist is:
τ_t = (42 kg)·(9.81 m/s2)·(1 m/2) = 206.91 N·m
The restoring torque due to the displacement of the trapeze artist from the equilibrium position is:
τ_r = (48 kg)·(9.81 m/s2)·(1 m/2)·θ = 24.405θ N·m
where θ is the angular displacement of the system from the equilibrium position.
At the natural frequency, the restoring torque and the torque due to the weight of the system balance each other:
τ_r = τ_l + τ_t
Substituting the values, we get:
24.405θ = 58.86 + 206.91
24.405θ = 265.77
θ = 10.88 radians
The angular frequency is:
ω = √(g/l) = √(9.81 m/s2/1 m) = 3.13 rad/s (approx.)
This is close to the option B) 3.07 rad/s.

Question 3

An automobile (m = 1.00 × 10³ kg)is driven into a brick wall in a safety test.The bumper behaves like a spring (k = 5.00 × 10⁶ N/m),and is observed to compress a distance of 3.16 cm as the car is brought to rest.What was the initial speed of the automobile?

Accepted Answer

The answer of An automobile (m = 1.00 × 10³...

Question 4

An ore car of mass 4 000 kg starts from rest and rolls downhill on tracks from a mine.A spring with k = 400 000 N/m is located at the end of the tracks.At the spring's maximum compression,the car is at an elevation 10 m lower than its elevation at the starting point.How much is the spring compressed in stopping the ore car? Ignore friction.

Accepted Answer

The answer of An ore car of mass 4 000...

Question 5

Exhibit 15-1
A graph of position versus time for an object oscillating at the free end of a horizontal spring is shown below.A point or points at which the object has positive velocity and zero acceleration is(are)  Use this exhibit to answer the following question(s).
-Refer to Exhibit 15-1.A point or points at which the object has positive velocity and zero acceleration is(are)
A)B
B)C
C)D
D)B and D
E)A and E

Accepted Answer

The answer of Exhibit 15-1
A graph of position versus time...

Question 6

The mat of a trampoline is held by 32 springs,each having a spring constant of 5 000 N/m.A person with a mass of 40.0 kg jumps from a platform 1.93 m high onto the trampoline.Determine the stretch of each of the springs.

Accepted Answer

The answer of The mat of a trampoline is held...

Question 7

Simple harmonic oscillations can be modeled by the projection of circular motion at constant angular velocity onto a diameter of the circle.When this is done,the analog along the diameter of the acceleration of the particle executing simple harmonic motion is
A)the displacement from the center of the diameter of the projection of the position of the particle on the circle.
B)the projection along the diameter of the velocity of the particle on the circle.
C)the projection along the diameter of tangential acceleration of the particle on the circle.
D)the projection along the diameter of centripetal acceleration of the particle on the circle.
E)meaningful only when the particle moving in the circle also has a non-zero tangential acceleration.

Accepted Answer

The acceleration in simple harmonic motion is given by a = -ω^2x, where x is the displacement from equilibrium and ω is the angular frequency. The projection of circular motion onto a diameter results in a sinusoidal displacement along the diameter. The only acceleration that can be projected along this diameter is the centripetal acceleration, which is always perpendicular to the velocity and directed towards the center of the circle, and is given by a = ω^2r, where r is the radius of the circle. Since the projection of the position on the circle results in a sinusoidal displacement along the diameter, and the tangential acceleration is always perpendicular to the diameter (and hence cannot be projected onto it), choice D is the correct answer.

Question 8

A damped oscillator is released from rest with an initial displacement of 10.00 cm.At the end of the first complete oscillation the displacement reaches 9.05 cm.When 4 more oscillations are completed,what is the displacement reached?
A)7.41 cm
B)6.71 cm
C)6.07 cm
D)5.49 cm
E)5.25 cm

Accepted Answer

The amplitude of a damped oscillator decreases exponentially according to the equation $A=A_0e^{-\beta t}$, where $A_0$ is the initial amplitude, $\beta$ is the damping constant, and $t$ is time.

We can use the given information to solve for the damping constant:

$9.05 = 10e^{-\beta T}$, where T is the period of one oscillation.

Dividing by 10 and taking the natural logarithm of both sides, we get:

$-\beta T = \ln(0.905)$

$\beta = \frac{\ln(10/9.05)}{T} \approx 0.046 T^{-1}$

Next, we can use the damping constant to find the amplitude at the end of five oscillations:

$A = A_0e^{-5\beta T} = 10e^{-0.23T} \approx 6.07$ cm.

Therefore, the answer is C) 6.07 cm.

Question 9

Exhibit 15-1
A graph of position versus time for an object oscillating at the free end of a horizontal spring is shown below.A point or points at which the object has positive velocity and zero acceleration is(are)  Use this exhibit to answer the following question(s).
-Refer to Exhibit 15-1.The point at which the object has zero velocity and negative acceleration is
A)A
B)B
C)C
D)D
E)E

Accepted Answer

At point B, the velocity of the object is zero (as the line crosses the x-axis), and the acceleration is negative (as the slope of the line is negative, indicating a decrease in velocity with time).

Question 10

An archer pulls her bow string back 0.40 m by exerting a force that increases uniformly from zero to 240 N.What is the equivalent spring constant of the bow,and how much work is done in pulling the bow?

Accepted Answer

The answer of An archer pulls her bow string back...

Question 11

At sea level,at a latitude where ,a pendulum that takes 2.00 s for a complete swing back and forth has a length of 0.993 m.What is the value of g in m/s² at a location where the length of such a pendulum is 0.970 m? A)0.098 3 B)3.05 C)9.57 D)10.0 E)38.3

Accepted Answer

We can use the equation for the period of a simple pendulum:

T = 2π√(L/g)

where T is the period, L is the length of the pendulum and g is the acceleration due to gravity.

At the first location, we have:

T1 = 2.00 s 
L1 = 0.993 m

Solving for g1:

g1 = (4π^2 * L1) / T1^2 = 9.81 m/s^2

At the second location, we have:

T2 = 2.00 s 
L2 = 0.970 m

Solving for g2:

g2 = (4π^2 * L2) / T2^2 = 9.57 m/s^2

Therefore, the answer is C) 9.57 m/s^2.

Question 12

When a damping force is applied to a simple harmonic oscillator which has angular frequency ω₀ in the absence of damping,the new angular frequency ω is such that A)ω < ω₀. B)ω = ω₀. C)ω > ω₀. D)ωT < ω₀T₀. E)ωT > ω₀T₀.

Accepted Answer

When a damping force is applied to a simple harmonic oscillator, the amplitude of the oscillations decreases with time. This means that the effective angular frequency decreases as well. Therefore, the new angular frequency ω must be less than ω₀ in the absence of damping, making choice A the correct answer.

Question 13

In an inertia balance,a body supported against gravity executes simple harmonic oscillations in a horizontal plane under the action of a set of springs.If a 1.00 kg body vibrates at 1.00 Hz,a 2.00 kg body will vibrate at
A)0.500 Hz.
B)0.707 Hz.
C)1.00 Hz.
D)1.41 Hz.
E)2.00 Hz.

Accepted Answer

The frequency of vibration in an inertia balance depends on the square root of the ratio of the force constant of the springs to the mass of the body. That is, 
frequency = (1/2π) √(k/m). Since the force constant k remains the same for both bodies, and the mass of the 2.00 kg body is twice that of the 1.00 kg body, the frequency of vibration for the 2.00 kg body will be (1/2π) √(k/2.00) = (1/2) [(1/2π) √(k/1.00)]. Thus, the frequency of vibration for the 2.00 kg body will be 0.707 times the frequency of vibration for the 1.00 kg body, or 0.707 Hz.

Question 14

To double the total energy of a mass oscillating at the end of a spring with amplitude A,we need to
A)increase the angular frequency by  .
B)increase the amplitude by  .
C)increase the amplitude by 2.
D)increase the angular frequency by 2.
E)increase the amplitude by 4 and decrease the angular frequency by  .

Accepted Answer

The answer of To double the total energy of a...

Question 15

John says that the value of the function cos[ω(t + T)+ ϕ],obtained one period T after time t,is greater than cos(ωt + ϕ)by 2π.Larry says that it is greater by the addition of 1.00 to cos(ωt + ϕ).Which one,if either,is correct?
A)John,because ωT = 2π.
B)John,because ωT = 1 radian.
C)Larry,because ωT = 2π.
D)Larry,because ωT = 1 radian.
E)Neither,because cos(θ + 2π)= cosθ.

Accepted Answer

Neither John nor Larry is correct because the cosine function is periodic with a period of 2π, meaning that cos(θ + 2π) = cosθ. Therefore, the value of the function after one period T does not change by 2π or by the addition of 1.00; it remains the same.

Question 16

Which of the following combinations of variables results in the greatest period for a pendulum?
A)length = L,mass = M,and maximum angular displacement = 3 degrees
B)length = 2L,mass = M/2,and maximum angular displacement = 1 degree
C)length = 1.5L,mass = 2M,and maximum angular displacement = 2 degrees
D)length =  L,mass =  M,and maximum angular displacement =  degrees
E)length =  L,mass = 4M,and maximum angular displacement = 4 degrees

Accepted Answer

The period of a pendulum is given by T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. The period does not depend on the mass or maximum angular displacement of the pendulum. Therefore, option B with a length of 2L has the greatest period because the period is proportional to the square root of the length.

Question 17

The oscillation of the 2.0-kg mass on a spring is described by 
Where x is in centimeters and t is in seconds.What is the force constant of the spring?
A)4.0 N/m
B)0.80 N/m
C)16 N/m
D)32 N/m
E)2.0  N/m

Accepted Answer

The answer of The oscillation of the 2.0-kg mass on...

Question 18

When a damping force is applied to a simple harmonic oscillator which has period T₀ in the absence of damping,the new period T is such that A)T < T₀. B)T = T₀. C)T > T₀. D)ωT < ω₀T₀. E)ωT > ω₀T₀.

Accepted Answer

The answer of When a damping force is applied to...

Question 19

Ellen says that whenever the acceleration is directly proportional to the displacement of an object from its equilibrium position,the motion of the object is simple harmonic motion.Mary says this is true only if the acceleration is opposite in direction to the displacement.Which one,if either,is correct? A)Ellen,because ω² is directly proportional to the constant multiplying the displacement and to the mass. B)Ellen,because ω² is directly proportional to the mass. C)Mary,because ω² is directly proportional to the constant multiplying the displacement and to the mass. D)Mary,because ω² is directly proportional to the mass. E)Mary,because the second derivative of an oscillatory function like sin(ωt)or cos(ωt)is always proportional to the negative of the original function.

Accepted Answer

Mary is correct because for simple harmonic motion, the acceleration (which is the second derivative of displacement with respect to time) must be directly proportional to the displacement and in the opposite direction. This is mathematically represented in the behavior of functions like sin(ωt) or cos(ωt), where the second derivative is indeed proportional to the negative of the original function, indicating the acceleration is opposite in direction to the displacement.

Question 20

Exhibit 15-1
A graph of position versus time for an object oscillating at the free end of a horizontal spring is shown below.A point or points at which the object has positive velocity and zero acceleration is(are)  Use this exhibit to answer the following question(s).
-Refer to Exhibit 15-1.The point at which the object has zero velocity and positive acceleration is
A)A
B)B
C)C
D)D
E)E

Accepted Answer

The answer of Exhibit 15-1
A graph of position versus time...

Quiz 15: Oscillatory Motion

An automobile (m = 1.00 × 10³ kg)is driven into a brick wall in a safety test.The bumper behaves like a spring (k = 5.00 × 10⁶ N/m),and is observed to compress a distance of 3.16 cm as the car is brought to rest.What was the initial speed of the automobile?

The mat of a trampoline is held by 32 springs,each having a spring constant of 5 000 N/m.A person with a mass of 40.0 kg jumps from a platform 1.93 m high onto the trampoline.Determine the stretch of each of the springs.

Simple harmonic oscillations can be modeled by the projection of circular motion at constant angular velocity onto a diameter of the circle.When this is done,the analog along the diameter of the acceleration of the particle executing simple harmonic motion is

A damped oscillator is released from rest with an initial displacement of 10.00 cm.At the end of the first complete oscillation the displacement reaches 9.05 cm.When 4 more oscillations are completed,what is the displacement reached?

An archer pulls her bow string back 0.40 m by exerting a force that increases uniformly from zero to 240 N.What is the equivalent spring constant of the bow,and how much work is done in pulling the bow?

At sea level,at a latitude where ,a pendulum that takes 2.00 s for a complete swing back and forth has a length of 0.993 m.What is the value of g in m/s² at a location where the length of such a pendulum is 0.970 m?

When a damping force is applied to a simple harmonic oscillator which has angular frequency ω₀ in the absence of damping,the new angular frequency ω is such that

In an inertia balance,a body supported against gravity executes simple harmonic oscillations in a horizontal plane under the action of a set of springs.If a 1.00 kg body vibrates at 1.00 Hz,a 2.00 kg body will vibrate at

To double the total energy of a mass oscillating at the end of a spring with amplitude A,we need to

John says that the value of the function cos[ω(t + T) + ϕ],obtained one period T after time t,is greater than cos(ωt + ϕ) by 2π.Larry says that it is greater by the addition of 1.00 to cos(ωt + ϕ) .Which one,if either,is correct?

Which of the following combinations of variables results in the greatest period for a pendulum?

The oscillation of the 2.0-kg mass on a spring is described by Where x is in centimeters and t is in seconds.What is the force constant of the spring?

When a damping force is applied to a simple harmonic oscillator which has period T₀ in the absence of damping,the new period T is such that

Quiz 15: Oscillatory Motion

An automobile (m = 1.00 × 103 kg)is driven into a brick wall in a safety test.The bumper behaves like a spring (k = 5.00 × 106 N/m),and is observed to compress a distance of 3.16 cm as the car is brought to rest.What was the initial speed of the automobile?

The mat of a trampoline is held by 32 springs,each having a spring constant of 5 000 N/m.A person with a mass of 40.0 kg jumps from a platform 1.93 m high onto the trampoline.Determine the stretch of each of the springs.

Simple harmonic oscillations can be modeled by the projection of circular motion at constant angular velocity onto a diameter of the circle.When this is done,the analog along the diameter of the acceleration of the particle executing simple harmonic motion is

A damped oscillator is released from rest with an initial displacement of 10.00 cm.At the end of the first complete oscillation the displacement reaches 9.05 cm.When 4 more oscillations are completed,what is the displacement reached?

An archer pulls her bow string back 0.40 m by exerting a force that increases uniformly from zero to 240 N.What is the equivalent spring constant of the bow,and how much work is done in pulling the bow?

At sea level,at a latitude where ,a pendulum that takes 2.00 s for a complete swing back and forth has a length of 0.993 m.What is the value of g in m/s2 at a location where the length of such a pendulum is 0.970 m?

When a damping force is applied to a simple harmonic oscillator which has angular frequency ω0 in the absence of damping,the new angular frequency ω is such that

In an inertia balance,a body supported against gravity executes simple harmonic oscillations in a horizontal plane under the action of a set of springs.If a 1.00 kg body vibrates at 1.00 Hz,a 2.00 kg body will vibrate at

To double the total energy of a mass oscillating at the end of a spring with amplitude A,we need to

John says that the value of the function cos[ω(t + T) + ϕ],obtained one period T after time t,is greater than cos(ωt + ϕ) by 2π.Larry says that it is greater by the addition of 1.00 to cos(ωt + ϕ) .Which one,if either,is correct?

Which of the following combinations of variables results in the greatest period for a pendulum?

The oscillation of the 2.0-kg mass on a spring is described by Where x is in centimeters and t is in seconds.What is the force constant of the spring?

When a damping force is applied to a simple harmonic oscillator which has period T0 in the absence of damping,the new period T is such that

An automobile (m = 1.00 × 10³ kg)is driven into a brick wall in a safety test.The bumper behaves like a spring (k = 5.00 × 10⁶ N/m),and is observed to compress a distance of 3.16 cm as the car is brought to rest.What was the initial speed of the automobile?

At sea level,at a latitude where ,a pendulum that takes 2.00 s for a complete swing back and forth has a length of 0.993 m.What is the value of g in m/s² at a location where the length of such a pendulum is 0.970 m?

When a damping force is applied to a simple harmonic oscillator which has angular frequency ω₀ in the absence of damping,the new angular frequency ω is such that

When a damping force is applied to a simple harmonic oscillator which has period T₀ in the absence of damping,the new period T is such that