Question 1

A leaky faucet drips 40 times in 
What is the frequency of the dripping?
A)1.3 Hz
B)0.75 Hz
C)1.6 Hz
D)0.63 Hz

Accepted Answer

The answer of A leaky faucet drips 40 times in...

Question 2

An object undergoing simple harmonic motion has a maximum displacement of 
At 
If the angular frequency of oscillation is 
What is the object's displacement when 
A)4.8 m
B)5.6 m
C)3.7 m
D)3.1 m

Accepted Answer

The answer of An object undergoing simple harmonic motion has...

Question 3

An air-track cart is attached to a spring and completes one oscillation every 5.67 s in simple harmonic motion.At time t = 0.00 s the cart is released at the position x = +0.250 m.What is the position of the cart when t = 29.6 s?
A)x = 0.0461 m
B)x = 0.210 m
C)x = 0.218 m
D)x = 0.342 m
E)x = -0.218 m

Accepted Answer

The answer of An air-track cart is attached to a...

Question 4

A package is oscillating on a spring scale with a period of 4.60 s.At time t = 0.00 s the package has zero speed and is at x = 8.30 cm.At what time after t = 0.00 s will the package first be at x = 4.15 cm?
A)0.575 s
B)0.767 s
C)1.15 s
D)1.30 s
E)1.53 s

Accepted Answer

The motion of the package can be described by simple harmonic motion (SHM), where the displacement x from the equilibrium position as a function of time t is given by $x(t) = A \cos(\omega t + \phi)$, where A is the amplitude, $\omega$ is the angular frequency, and $\phi$ is the phase constant. At $t = 0$, the package is at its maximum displacement (amplitude $A = 8.30$ cm), and since it has zero speed, this corresponds to the cosine function starting from its maximum value. Therefore, the phase constant $\phi = 0$, and the equation simplifies to $x(t) = A \cos(\omega t)$.Given that the period $T = 4.60$ s, we can find the angular frequency $\omega$ using the relation $\omega = \frac{2\pi}{T}$. The displacement $x = 4.15$ cm is half the amplitude, which corresponds to the cosine function being equal to 0.5 (since the full amplitude represents cosine of 0 or $\pi$, and half amplitude corresponds to $\cos^{-1}(0.5)$).To find the time when $x = 4.15$ cm, we solve $4.15 = 8.30 \cos(\omega t)$ or $\cos(\omega t) = 0.5$. This gives $\omega t = \frac{\pi}{3}$ or $\frac{5\pi}{3}$ for the first instance within the period, but since we start from maximum displacement and move towards the equilibrium (decreasing x), we use $\omega t = \frac{\pi}{3}$.Substituting $\omega = \frac{2\pi}{4.60}$ into $\omega t = \frac{\pi}{3}$ gives $t = \frac{T}{6}$. Calculating this gives $t = \frac{4.60}{6} = 0.767$ s, which is the first time the package will be at $x = 4.15$ cm after $t = 0.00$ s.

Question 5

If your heart is beating at 76.0 beats per minute,what is the frequency of your heart's oscillations in hertz?
A)4560 Hz
B)1450 Hz
C)3.98 Hz
D)2.54 Hz
E)1.27 Hz

Accepted Answer

The frequency in hertz (Hz) is the number of cycles (or beats) per second. To convert beats per minute to hertz, divide the number of beats by 60 seconds. Therefore, 76.0 beats per minute is 76.0/60 = 1.27 Hz.

Question 6

The position of an air-track cart that is oscillating on a spring is given by the equation x = (12.4 cm)cos[(6.35 s^-1)t].At what value of t after t = 0.00 s is the cart first located at x = 8.47 cm? A)4.34 s B)0.108 s C)0.129 s D)7.39 s E)7.75 s

Accepted Answer

To find the time when the cart is first located at x=8.47 cm, we need to solve for t in the given equation x = (12.4 cm)cos[(6.35 s^-1)t] = 8.47 cm
cos[(6.35 s^-1)t] = 8.47 cm / (12.4 cm) = 0.683
Taking the inverse cosine of both sides, we have:
(6.35 s^-1)t = cos^-1(0.683) = 0.819 radians
t = 0.819 radians / (6.35 s^-1) = 0.129 s
Therefore, the answer is (C) 0.129 s.

Question 7

An object is undergoing simple harmonic motion of amplitude 2.3 m.If the maximum velocity of the object is 10 m/s,what is the object's angular frequency?
A)4.3 rad/s
B)4.8 rad/s
C)3.5 rad/s
D)4.0 rad/s

Accepted Answer

The maximum velocity $v_{max}$ in simple harmonic motion is given by $v_{max} = \omega A$, where $\omega$ is the angular frequency and $A$ is the amplitude. Rearranging for $\omega$, we get $\omega = \frac{v_{max}}{A} = \frac{10 \, \text{m/s}}{2.3 \, \text{m}} \approx 4.3 \, \text{rad/s}$.

Question 8

A pendulum of length L is suspended from the ceiling of an elevator.When the elevator is at rest the period of the pendulum is T.How would the period of the pendulum change if the supporting chain were to break,putting the elevator into freefall?
A)The period does not change.
B)The period increases slightly.
C)The period decreases slightly.
D)The period becomes zero.
E)The period becomes infinite because the pendulum would not swing.

Accepted Answer

When the elevator is in freefall, both the pendulum and the elevator are in a state of freefall, experiencing the same acceleration due to gravity. This means there is no force to restore the pendulum to its equilibrium position, so it would not swing at all. Hence, the concept of a period for the pendulum's motion does not apply, effectively making it infinite because the pendulum does not initiate any oscillatory motion.

Question 9

A pendulum of length L is suspended from the ceiling of an elevator.When the elevator is at rest the period of the pendulum is T.How does the period of the pendulum change when the elevator moves upward with constant acceleration?
A)The period does not change.
B)The period increases.
C)The period decreases.
D)The period becomes zero.
E)The period increases if the upward acceleration is more than g/2 but decreases if the upward acceleration is less than g/2.

Accepted Answer

The period of a simple pendulum is given by T = 2π √(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. When the elevator moves upward with constant acceleration a, the effective acceleration acting on the pendulum is g + a.

Using this effective acceleration in the formula for the period of the pendulum, we get:

T' = 2π √(L/(g+a))

Since g + a is larger than g, the square root term in the above formula is smaller than the square root term in the original formula for T. Therefore, T' is smaller than T, which means the period of the pendulum decreases when the elevator moves upward with constant acceleration.

Therefore, the correct choice is C.

Question 10

The quartz crystal in a digital watch has a frequency of 32.8 kHz.What is its period of oscillation?
A)30.5 µs
B)15.3 µs
C)95.8 µs
D)0.191 ms
E)9.71 µs

Accepted Answer

The period of oscillation can be calculated using the formula: T = 1/frequency
T = 1/32.8 kHz = 30.5 µs (as 1 kHz = 1000 Hz)

Question 11

A simple pendulum and a mass oscillating on an ideal spring both have period T in an elevator at rest.If the elevator now accelerates downward uniformly at 2 m/s²,what is true about the periods of these two systems? A)Both periods would remain the same. B)Both periods would increase. C)Both periods would decrease. D)The period of the pendulum would increase but the period of the spring would stay the same. E)The period of the pendulum would decrease but the period of the spring would stay the same.

Accepted Answer

The period of a simple pendulum depends on the acceleration due to gravity (g). If the elevator accelerates downward, the effective acceleration due to gravity decreases, leading to an increase in the pendulum's period. The period of a mass on a spring depends on the mass and the spring constant, not on gravity, so it remains unchanged.

Question 12

A guitar string is set into vibration with a frequency of 512 Hz.How many oscillations does it undergo each minute?
A)30,700
B)8.53
C)26.8
D)1610
E)512

Accepted Answer

To find the number of oscillations per minute, we need to convert the frequency from Hz to oscillations per minute. There are 60 seconds in one minute, so in one second, the guitar string oscillates 512 times.

Frequency = number of oscillations per second

Frequency = 512 Hz

Number of oscillations in one minute = 512 Hz x 60 seconds = 30,720 oscillations per minute

However, we need to be careful when rounding our answer. The closest option is A) 30,700.

Question 13

A pendulum of length L is suspended from the ceiling of an elevator.When the elevator is at rest the period of the pendulum is T.How does the period of the pendulum change when the elevator moves downward with constant acceleration?
A)The period does not change.
B)The period increases.
C)The period decreases.
D)The period becomes zero.
E)The period increases if the upward acceleration is more than g/2 but decreases if the upward acceleration is less than g/2.

Accepted Answer

When the elevator moves downward with constant acceleration, the effective gravitational acceleration decreases, leading to an increase in the period of the pendulum.

Question 14

If the frequency of a system undergoing simple harmonic motion doubles,by what factor does the maximum value of acceleration change?
A)4
B)2
C) 
D)2/π

Accepted Answer

The answer of If the frequency of a system undergoing...

Question 15

Grandfather clocks are designed so they can be adjusted by moving the weight at the bottom of the pendulum up or down.Suppose you have a grandfather clock at home that runs fast.Which of the following adjustments of the weight would make it more accurate? (There could be more than one correct choice. )
A)Raise the weight.
B)Lower the weight.
C)Add more mass to the weight.
D)Remove some mass from the weight.
E)Decrease the amplitude of swing by a small amount.

Accepted Answer

The time it takes for the pendulum to swing back and forth is determined by the length of the pendulum. By lowering the weight, the effective length of the pendulum is increased, making the clock run more slowly and therefore more accurate. Raising the weight would do the opposite and make the clock run even faster. Adding or removing mass from the weight would have no effect on the clock's accuracy. Decreasing the amplitude of the swing would also have little effect on accuracy, as long as the swing is still wide enough for the escapement mechanism to function properly.

Question 16

A simple pendulum that consists of a small ball of mass m and a massless wire of length L swings with a period T.Suppose now that the mass is rearranged so that mass of the ball was reduced but the mass of the wire was increased,with the total mass remaining m and the length being L.What is true about the new period of swing? (There could be more than one correct choice. )
A)The new period is T because the total mass m has not changed.
B)The new period is T because the length L has not changed.
C)The new period is greater than T.
D)The new period is less than T
E)The new period is T because neither L nor m have changed.

Accepted Answer

The period T of a simple pendulum is given by T=2π√(L/g), where g is the acceleration due to gravity. As the mass of the ball is reduced and the mass of the wire is increased, the moment of inertia of the pendulum increases (since the mass is farther from the axis of rotation), making it take longer to swing back and forth. Therefore, the new period of swing is longer than T, and the correct choice is D.

Question 17

The position of an object that is oscillating on a spring is given by the equation x = (17.4 cm)cos[(5.46 s^-1)t].What is the angular frequency for this motion? A)0.183 rad/s B)5.46 rad/s C)2.34 rad/s D)17.4 rad/s E)0.869 rad/s

Accepted Answer

In the given equation x = (17.4 cm)cos[(5.46 rad/s)t], the angular frequency is the coefficient of 't' inside the cosine function, which is 5.46 rad/s. Therefore, the correct option is B.

Question 18

A simple pendulum and a mass oscillating on an ideal spring both have period T in an elevator at rest.If the elevator now moves downward at a uniform 2 m/s,what is true about the periods of these two systems?
A)Both periods would remain the same.
B)Both periods would increase.
C)Both periods would decrease.
D)The period of the pendulum would increase but the period of the spring would stay the same.
E)The period of the pendulum would decrease but the period of the spring would stay the same.

Accepted Answer

The period of a simple pendulum depends on the length of the pendulum and the acceleration due to gravity, neither of which changes in an elevator moving downward at a constant speed. The period of a mass oscillating on an ideal spring depends on the mass and the spring constant, which also do not change in this scenario. Therefore, both periods remain the same.

Question 19

An object is oscillating on a spring with a period of 4.60 s.At time t = 0.00 s the object has zero speed and is at x = 8.30 cm.What is the acceleration of the object at t = 2.50 s? A)1.33 cm/s² B)0.784 cm/s² C)11.5 cm/s² D)14.9 cm/s² E)0.00 cm/s²

Accepted Answer

The answer of An object is oscillating on a spring...

Question 20

A pendulum of length L is suspended from the ceiling of an elevator.When the elevator is at rest the period of the pendulum is T.How does the period of the pendulum change when the elevator moves upward with constant velocity?
A)The period does not change.
B)The period increases.
C)The period decreases.
D)The period becomes zero.
E)The period increases if the upward acceleration is more than g/2 but decreases if the upward acceleration is less than g/2.

Accepted Answer

The period of a pendulum is determined by the length of the pendulum and the acceleration due to gravity. In this case, as the elevator moves upward with a constant velocity, the acceleration experienced by the pendulum remains the same as that of gravity. Thus, the period of the pendulum remains unchanged.

Quiz 13: Oscillations About Equilibrium

A leaky faucet drips 40 times in What is the frequency of the dripping?

An object undergoing simple harmonic motion has a maximum displacement of At If the angular frequency of oscillation is What is the object's displacement when

An air-track cart is attached to a spring and completes one oscillation every 5.67 s in simple harmonic motion.At time t = 0.00 s the cart is released at the position x = +0.250 m.What is the position of the cart when t = 29.6 s?

A package is oscillating on a spring scale with a period of 4.60 s.At time t = 0.00 s the package has zero speed and is at x = 8.30 cm.At what time after t = 0.00 s will the package first be at x = 4.15 cm?

If your heart is beating at 76.0 beats per minute,what is the frequency of your heart's oscillations in hertz?

The position of an air-track cart that is oscillating on a spring is given by the equation x = (12.4 cm) cos[(6.35 s^-1) t].At what value of t after t = 0.00 s is the cart first located at x = 8.47 cm?

An object is undergoing simple harmonic motion of amplitude 2.3 m.If the maximum velocity of the object is 10 m/s,what is the object's angular frequency?

A pendulum of length L is suspended from the ceiling of an elevator.When the elevator is at rest the period of the pendulum is T.How would the period of the pendulum change if the supporting chain were to break,putting the elevator into freefall?

A pendulum of length L is suspended from the ceiling of an elevator.When the elevator is at rest the period of the pendulum is T.How does the period of the pendulum change when the elevator moves upward with constant acceleration?

The quartz crystal in a digital watch has a frequency of 32.8 kHz.What is its period of oscillation?

A simple pendulum and a mass oscillating on an ideal spring both have period T in an elevator at rest.If the elevator now accelerates downward uniformly at 2 m/s²,what is true about the periods of these two systems?

A guitar string is set into vibration with a frequency of 512 Hz.How many oscillations does it undergo each minute?

A pendulum of length L is suspended from the ceiling of an elevator.When the elevator is at rest the period of the pendulum is T.How does the period of the pendulum change when the elevator moves downward with constant acceleration?

If the frequency of a system undergoing simple harmonic motion doubles,by what factor does the maximum value of acceleration change?

The position of an object that is oscillating on a spring is given by the equation x = (17.4 cm) cos[(5.46 s^-1) t].What is the angular frequency for this motion?

A simple pendulum and a mass oscillating on an ideal spring both have period T in an elevator at rest.If the elevator now moves downward at a uniform 2 m/s,what is true about the periods of these two systems?

An object is oscillating on a spring with a period of 4.60 s.At time t = 0.00 s the object has zero speed and is at x = 8.30 cm.What is the acceleration of the object at t = 2.50 s?

A pendulum of length L is suspended from the ceiling of an elevator.When the elevator is at rest the period of the pendulum is T.How does the period of the pendulum change when the elevator moves upward with constant velocity?