Question 1

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 presence of a damping force in a simple harmonic oscillator increases the period, making T greater than T₀.

Question 2

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. Therefore, to increase the period, we want to increase the length and/or decrease the value of g. Option B has the longest length (2L), and since the mass is halved, g is also halved (since g = GM/R^2). Therefore, option B results in the greatest period.

Question 3

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...

Question 4

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 a simple harmonic oscillator depends on the mass and the spring constant. The formula for the frequency is given by:
f = 1/(2π)√(k/m)
where f is the frequency, k is the spring constant, and m is the mass. 
Since the spring constant and the support remain the same, we can equate the formula for the two masses:
f1 = 1/(2π)√(k/1)
f2 = 1/(2π)√(k/2)
Dividing these two equations gives:
f2/f1 = √(k/2)/√(k/1) 
f2/f1 = √2 
f2 = f1√2 
Substituting f1 = 1 Hz, we get:
f2 = 1 Hz x √2 ≈ 0.707 Hz 
Therefore, the 2.00 kg body will vibrate at approximately 0.707 Hz or choice B.

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

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 total energy of a mass oscillating at the end of a spring is given by E = (1/2)kA^2 where k is the spring constant and A is the amplitude of oscillation. To double the total energy, we need to increase A by a factor of sqrt(2) (since energy is directly proportional to the square of amplitude) and keep the angular frequency the same. Therefore, the correct choice is B which suggests increasing the amplitude by 11eb6adc_ee8e_13d5_b0cf_a726a3b3edc5_TB8245_11.

Question 7

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

We can use the equation for the amplitude of a damped oscillator:
A = A0 e^(-βt)
where A is the amplitude at time t, A0 is the initial amplitude, β is the damping coefficient.

From the statement of the problem: 
- A0 = 10.00 cm, 
- A = 9.05 cm after the first complete oscillation (T is the period of oscillation): 
9.05 = 10.00 e^(-βT), 
- we want to find A after 5 more oscillations: 
A = A0 e^(-5βT).

We can solve for βT by taking the natural logarithm of both sides of the equation for the first complete oscillation: 
ln(9.05) = ln(10.00) - βT, 
βT = ln(10.00) - ln(9.05) ≈ 0.0996 s^-1.

Substituting this into the equation for A after 5 more oscillations: 
A = 10.00 e^(-5βT) = 10.00 e^(-0.498 s^-1) ≈ 6.07 cm.

Therefore, the answer is C.

Question 8

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 9

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 10

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 system's natural frequency can be calculated using the formula for the angular frequency of a physical pendulum: ω = sqrt(g / L_eff), where L_eff is the effective length of the pendulum. Here, L_eff is the distance from the pivot to the center of mass of the system. The center of mass is at the midpoint of the ladder, so L_eff = 1.00 m. Using g = 9.81 m/s², ω = sqrt(9.81 / 1.00) ≈ 3.13 s⁻¹, which is closest to 3.07 s⁻¹.

Question 11

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 projection of circular motion onto a diameter of the circle results in simple harmonic motion. The acceleration of the particle at any point on the circle has two components - tangential acceleration and centripetal acceleration. When the circular motion is projected onto a diameter, the tangential acceleration becomes zero as there is no component of the velocity vector along the diameter. Therefore, the analog of acceleration along the diameter of the projection of circular motion is the projection along the diameter of the centripetal acceleration, which is towards the center of the circle. This is because the centripetal acceleration is responsible for keeping the particle in circular motion and towards the center of the circle.

Question 12

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

Using the equation T=2π√(L/g), we can solve for g at sea level: 
2.00=sqrt[(0.993)/g] * 2π 
g=9.81 m/s^2

We can then use this value of g to solve for the period of the pendulum at the new location: 
T=2π√(L/g) 
T=2π√(0.970/9.81) 
T=1.97 s

Finally, we can use this period to solve for the value of g at the new location: 
1.97=sqrt[(0.970)/g] * 2π 
g=9.57 m/s^2

Therefore, the answer is C.

Question 13

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 14

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 15

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

At point C, the position of the object is zero (crossing the x-axis), which means it is passing through its equilibrium position. At this point, the velocity of the object is negative (moving towards the left) and the acceleration is zero (since it is at the maximum displacement from equilibrium).

Question 16

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 17

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, the energy of the oscillator is continuously dissipated, causing the amplitude to decrease over time. This means that the angular frequency of the oscillator decreases as well. Therefore, ω < ω₀.

Question 18

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 19

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 20

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...

Quiz 15: Oscillatory Motion

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

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

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

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?

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?

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

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?

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?

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.

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

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?

Quiz 15: Oscillatory Motion

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

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

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

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?

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?

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

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?

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?

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.

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

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?

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

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

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?