Question 1

When a 0.350-kg package is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The package is now displaced from its equilibrium position and undergoes simple harmonic oscillations when released. What is the period of the oscillations?
A) 0.695 s
B) 0.483 s
C) 0.286 s
D) 0.0769 s
E) 1.44 s

Accepted Answer

The period of oscillations for a mass-spring system is given by $T = 2\pi\sqrt{\frac{m}{k}}$, where $m$ is the mass and $k$ is the spring constant. First, we find $k$ using $F = kx$, where $F = mg$ and $x$ is the displacement (0.12 m). With $g = 9.8 \, \text{m/s}^2$, $k = \frac{mg}{x} = \frac{0.350 \times 9.8}{0.12}$. Then, plug $m$ and $k$ into the formula for $T$.

Question 2

The equation of motion of a particle undergoing simple harmonic motion in the y direction is y = (2.0 cm) sin(0.60 s^-1 t). At time t = 0.60 s determine the particle's (a) position, (b) velocity, and (c) acceleration.

Accepted Answer

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Question 3

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 4

A 0.150-kg air track cart is attached to an ideal spring with a force constant (spring constant) of 3.58 N/m and undergoes simple harmonic oscillations. What is the period of the oscillations?
A) 2.57 s
B) 0.527 s
C) 0.263 s
D) 1.14 s
E) 1.29 s

Accepted Answer

The answer of A 0.150-kg air track cart is attached...

Question 5

If the frequency of the motion of a simple harmonic oscillator is doubled, by what factor does the maximum speed of the oscillator change?
A) 2
B) 4
C) It does not change.
D) 1/2
E) 1/4

Accepted Answer

The maximum speed of the oscillator is directly proportional to the frequency, as given by the equation v = 2πfA, where v is the maximum speed, f is the frequency, and A is the amplitude of the oscillator. When the frequency is doubled, the maximum speed is also doubled, as all other variables remain constant. Therefore, the answer is A)2.

Question 6

A 0.25 kg harmonic oscillator has a total mechanical energy of $4.1 \mathrm {~J}$ If the oscillation amplitude is $20.0 \mathrm {~cm} ,$ what is the oscillation frequency?
A) 4.6 Hz
B) 1.4 Hz
C) 2.3 Hz
D) 3.2 Hz

Accepted Answer

We can use the equation for total mechanical energy of a harmonic oscillator: 
$$E = \frac{1}{2} k A^2$$
where $k$ is the spring constant and $A$ is the amplitude of oscillation.

We can solve for the spring constant: 
$$k = \frac{2E}{A^2} = \frac{2(4.1 \mathrm{~J})}{(0.20 \mathrm{~m})^2} \approx 102.5 \mathrm{~N/m}$$

Then, we can use the equation for the angular frequency of a simple harmonic oscillator: 
$$\omega = \sqrt{\frac{k}{m}}$$
where $m$ is the mass of the oscillator.

Plugging in the values, we get: 
$$\omega = \sqrt{\frac{102.5 \mathrm{~N/m}}{0.25 \mathrm{~kg}}} \approx 20.3 \mathrm{~rad/s}$$

Finally, we can convert the angular frequency to the oscillation frequency: 
$$f = \frac{\omega}{2\pi} \approx \frac{20.3 \mathrm{~rad/s}}{2\pi} \approx 3.23 \mathrm{~Hz} \approx \boxed{	extbf{(A) 4.6 Hz}}$$

Question 7

The position of an object that is oscillating on an ideal spring is given by x = (17.4 cm) cos[(5.46 s^-1)t]. Write an expression for the acceleration of the particle as a function of time using the cosine function.

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Question 8

The position of an object that is oscillating on a spring is given by the equation x = (18.3 cm) cos[(2.35 s^-1)t]. What are the (a) frequency, (b) amplitude, and (c) period of this motion?

Accepted Answer

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Question 9

A ball is oscillating on an ideal spring with an amplitude of 8.3 cm and a period of 4.6 s. Write an expression for its position, x, as a function of time t, if x is equal to 8.3 cm at t = 0.0 s. Use the cosine function.

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Question 10

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

The equation for the position of an object that is oscillating on a spring is x = Acos(ωt), where A is the amplitude and ω is the angular frequency. Comparing this to the given equation x = (17.4 cm)cos[(5.46s^-1)t], we can see that the angular frequency is equal to 5.46 rad/s. Therefore, the correct answer is choice B.

Question 11

A 0.250-kg stone is attached to an ideal spring and undergoes simple harmonic oscillations with a period of 0.640 s. What is the force constant (spring constant) of the spring?
A) 2.45 N/m
B) 12.1 N/m
C) 24.1 N/m
D) 0.102 N/m
E) 0.610 N/m

Accepted Answer

The period of a mass-spring system is given by: 
$T=2\pi\sqrt{\frac{m}{k}}$
where $m$ is the mass of the object and $k$ is the spring constant.
Rearranging the formula, we can solve for the spring constant:
$k=\frac{4\pi^2 m}{T^2}$
Substituting the given values, we get:
$k=\frac{4\pi^2 (0.250	ext{ kg})}{(0.640	ext{ s})^2}=24.1	ext{ N/m}$

Therefore, the answer is C.

Question 12

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 13

If the amplitude of the motion of a simple harmonic oscillator is doubled, by what factor does the maximum speed of the oscillator change?
A) 2
B) 4
C) It does not change.
D) 1/2
E) 1/4

Accepted Answer

The maximum speed $v_{max}$ of a simple harmonic oscillator is directly proportional to the amplitude $A$, as given by the equation $v_{max} = \omega A$, where $\omega$ is the angular frequency. Therefore, if the amplitude is doubled, the maximum speed also doubles.

Question 14

An object undergoing simple harmonic motion has a maximum displacement of $6.2 \mathrm {~m}$ at $t = 0.00 \mathrm {~s}$ If the angular frequency of oscillation is $1.6 \mathrm { rad } / \mathrm { s } \text {, }$ what is the object's displacement when $t = 3.5 \mathrm {~s} ?$
A) 4.8 m
B) 5.6 m
C) 3.7 m
D) 3.1 m

Accepted Answer

The correct answer is A.The object's displacement is given by the equation:$$x = A \cos(\omega t)$$where A is the maximum displacement, $\omega$ is the angular frequency, and t is the time.At t = 0.00 s, the object's displacement is 6.2 m, so A = 6.2 m.At t = 3.5 s, the object's displacement is:$$x = 6.2 \cos(1.6 	imes 3.5) = 4.8 \mathrm {~m}$$

Question 15

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 16

If the angular frequency of the motion of a simple harmonic oscillator is doubled, by what factor does the maximum acceleration of the oscillator change?
A) 2
B) 4
C) It does not change.
D) 1/2
E) 1/4

Accepted Answer

The maximum acceleration of a simple harmonic oscillator is proportional to the square of its angular frequency. Therefore, if the angular frequency is doubled, the maximum acceleration will be quadrupled (2^2 = 4). Hence, the correct answer is B.

Question 17

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

We need to solve for t in the equation x = (12.4 cm) cos[(6.35 s^-1)t] = 8.47 cm.
Setting up the equation and solving for t, we get:

(12.4 cm) cos[(6.35 s^-1)t] = 8.47 cm
cos[(6.35 s^-1)t] = 8.47/12.4
cos[(6.35 s^-1)t] = 0.683
(6.35 s^-1)t = cos^-1(0.683)
t = (cos^-1(0.683))/(6.35 s^-1)

Using a calculator, we find:
t ≈ 0.129 s

Therefore, the answer is (C) 0.129 s.

Question 18

In a supermarket, you place a 22.3-N (around 5 lb) bag of oranges on a scale, and the scale starts to oscillate at 2.7 Hz. What is the force constant (spring constant) of the spring of the scale?
A) 650 N/m
B) 600 N/m
C) 330 N/m
D) 820 N/m
E) 410 N/m

Accepted Answer

The force constant (spring constant) $k$ can be found using the formula for the period $T$ of a mass-spring system: $T = 2\pi\sqrt{\frac{m}{k}}$, where $m$ is the mass and $k$ is the spring constant. However, since we are given the frequency $f = 2.7 \, \text{Hz}$, we can use the relation $f = \frac{1}{T}$ to find $T$, and then solve for $k$. The mass $m$ can be found by dividing the weight of the bag by the acceleration due to gravity ($g = 9.8 \, \text{m/s}^2$). The weight of the bag is $22.3 \, \text{N}$, so $m = \frac{22.3 \, \text{N}}{9.8 \, \text{m/s}^2} \approx 2.27 \, \text{kg}$.Given $f = 2.7 \, \text{Hz}$, we have $T = \frac{1}{f} = \frac{1}{2.7} \approx 0.37 \, \text{s}$.Substituting $T$ back into the formula for $T$ in terms of $m$ and $k$, and solving for $k$, we get:$$0.37 = 2\pi\sqrt{\frac{2.27}{k}}$$Solving for $k$, we find $k \approx 650 \, \text{N/m}$.

Question 19

An object oscillates such that its position x as a function of time t obeys the equation x = (0.222 m) sin(314 s^-1 t), where t is in seconds. (a) In one period, what total distance does the object move? (b) What is the frequency of the motion? (c) What is the position of the object when t = 1.00 s?

Accepted Answer

The answer of An object oscillates such that its position...

Question 20

The position of an object that is oscillating on an ideal spring is given by x = (17.4 cm) cos[(5.46 s^-1)t]. Write an expression for the velocity of the particle as a function of time using the sine function.

Accepted Answer

The answer of The position of an object that is...

Quiz 14: Oscillations

When a 0.350-kg package is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The package is now displaced from its equilibrium position and undergoes simple harmonic oscillations when released. What is the period of the oscillations?

The equation of motion of a particle undergoing simple harmonic motion in the y direction is y = (2.0 cm) sin(0.60 s^-1 t). At time t = 0.60 s determine the particle's (a) position, (b) velocity, and (c) acceleration.

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 0.150-kg air track cart is attached to an ideal spring with a force constant (spring constant) of 3.58 N/m and undergoes simple harmonic oscillations. What is the period of the oscillations?

If the frequency of the motion of a simple harmonic oscillator is doubled, by what factor does the maximum speed of the oscillator change?

A 0.25 kg harmonic oscillator has a total mechanical energy of $4.1 \mathrm {~J}$ If the oscillation amplitude is $20.0 \mathrm {~cm} ,$ what is the oscillation frequency?

The position of an object that is oscillating on an ideal spring is given by x = (17.4 cm) cos[(5.46 s^-1)t]. Write an expression for the acceleration of the particle as a function of time using the cosine function.

The position of an object that is oscillating on a spring is given by the equation x = (18.3 cm) cos[(2.35 s^-1)t]. What are the (a) frequency, (b) amplitude, and (c) period of this motion?

A ball is oscillating on an ideal spring with an amplitude of 8.3 cm and a period of 4.6 s. Write an expression for its position, x, as a function of time t, if x is equal to 8.3 cm at t = 0.0 s. Use the cosine function.

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 0.250-kg stone is attached to an ideal spring and undergoes simple harmonic oscillations with a period of 0.640 s. What is the force constant (spring constant) of the spring?

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 the amplitude of the motion of a simple harmonic oscillator is doubled, by what factor does the maximum speed of the oscillator change?

An object undergoing simple harmonic motion has a maximum displacement of $6.2 \mathrm {~m}$ at $t = 0.00 \mathrm {~s}$ If the angular frequency of oscillation is $1.6 \mathrm { rad } / \mathrm { s } \text {, }$ what is the object's displacement when $t = 3.5 \mathrm {~s} ?$

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?

If the angular frequency of the motion of a simple harmonic oscillator is doubled, by what factor does the maximum acceleration of the oscillator change?

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?

In a supermarket, you place a 22.3-N (around 5 lb) bag of oranges on a scale, and the scale starts to oscillate at 2.7 Hz. What is the force constant (spring constant) of the spring of the scale?

An object oscillates such that its position x as a function of time t obeys the equation x = (0.222 m) sin(314 s^-1 t), where t is in seconds. (a) In one period, what total distance does the object move? (b) What is the frequency of the motion? (c) What is the position of the object when t = 1.00 s?

The position of an object that is oscillating on an ideal spring is given by x = (17.4 cm) cos[(5.46 s^-1)t]. Write an expression for the velocity of the particle as a function of time using the sine function.

Quiz 14: Oscillations

When a 0.350-kg package is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The package is now displaced from its equilibrium position and undergoes simple harmonic oscillations when released. What is the period of the oscillations?

The equation of motion of a particle undergoing simple harmonic motion in the y direction is y = (2.0 cm) sin(0.60 s-1 t). At time t = 0.60 s determine the particle's (a) position, (b) velocity, and (c) acceleration.

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 0.150-kg air track cart is attached to an ideal spring with a force constant (spring constant) of 3.58 N/m and undergoes simple harmonic oscillations. What is the period of the oscillations?

If the frequency of the motion of a simple harmonic oscillator is doubled, by what factor does the maximum speed of the oscillator change?

A 0.25 kg harmonic oscillator has a total mechanical energy of 4.1 J4.1 \mathrm {~J}4.1 J If the oscillation amplitude is 20.0 cm,20.0 \mathrm {~cm} ,20.0 cm, what is the oscillation frequency?

The position of an object that is oscillating on an ideal spring is given by x = (17.4 cm) cos[(5.46 s-1)t]. Write an expression for the acceleration of the particle as a function of time using the cosine function.

The position of an object that is oscillating on a spring is given by the equation x = (18.3 cm) cos[(2.35 s-1)t]. What are the (a) frequency, (b) amplitude, and (c) period of this motion?

A ball is oscillating on an ideal spring with an amplitude of 8.3 cm and a period of 4.6 s. Write an expression for its position, x, as a function of time t, if x is equal to 8.3 cm at t = 0.0 s. Use the cosine function.

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 0.250-kg stone is attached to an ideal spring and undergoes simple harmonic oscillations with a period of 0.640 s. What is the force constant (spring constant) of the spring?

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 the amplitude of the motion of a simple harmonic oscillator is doubled, by what factor does the maximum speed of the oscillator change?

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?

If the angular frequency of the motion of a simple harmonic oscillator is doubled, by what factor does the maximum acceleration of the oscillator change?

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?

In a supermarket, you place a 22.3-N (around 5 lb) bag of oranges on a scale, and the scale starts to oscillate at 2.7 Hz. What is the force constant (spring constant) of the spring of the scale?

An object oscillates such that its position x as a function of time t obeys the equation x = (0.222 m) sin(314 s-1 t), where t is in seconds. (a) In one period, what total distance does the object move? (b) What is the frequency of the motion? (c) What is the position of the object when t = 1.00 s?

The position of an object that is oscillating on an ideal spring is given by x = (17.4 cm) cos[(5.46 s-1)t]. Write an expression for the velocity of the particle as a function of time using the sine function.

The equation of motion of a particle undergoing simple harmonic motion in the y direction is y = (2.0 cm) sin(0.60 s^-1 t). At time t = 0.60 s determine the particle's (a) position, (b) velocity, and (c) acceleration.

A 0.25 kg harmonic oscillator has a total mechanical energy of $4.1 \mathrm {~J}$ If the oscillation amplitude is $20.0 \mathrm {~cm} ,$ what is the oscillation frequency?

The position of an object that is oscillating on an ideal spring is given by x = (17.4 cm) cos[(5.46 s^-1)t]. Write an expression for the acceleration of the particle as a function of time using the cosine function.

The position of an object that is oscillating on a spring is given by the equation x = (18.3 cm) cos[(2.35 s^-1)t]. What are the (a) frequency, (b) amplitude, and (c) period of this motion?

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?

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?

The position of an object that is oscillating on an ideal spring is given by x = (17.4 cm) cos[(5.46 s^-1)t]. Write an expression for the velocity of the particle as a function of time using the sine function.