Exam Review Guidance: Oscillatory Motion, Simple Harmonic Motion, Springs, Pendulums, and Energy

zQ1. A violin string playing the note A oscillates at 440 Hz. What’s its oscillation period?

Background

Topic: Oscillatory Motion – Frequency and Period

This question tests your understanding of the relationship between frequency and period in oscillatory motion.

Key Terms and Formulas

• Frequency ($f$): Number of oscillations per second, measured in Hertz (Hz).

• Period ($T$): Time for one complete oscillation, measured in seconds (s).

• Relationship: $T = \frac{1}{f}$

Step-by-Step Guidance

1. Identify the given frequency: $f = 440$ Hz.

2. Recall the formula relating period and frequency: $T = \frac{1}{f}$.

3. Set up the calculation: Substitute the value of $f$ into the formula.

Try solving on your own before revealing the answer!

Q2. The vibration frequency of a hydrogen chloride molecule is $8.66 \times 10^{13}$ Hz. How long does it take the molecule to complete one oscillation?

Background

Topic: Oscillatory Motion – Frequency and Period

This question is similar to Q1, testing your ability to convert frequency to period for molecular vibrations.

Key Terms and Formulas

• Frequency ($f$): $8.66 \times 10^{13}$ Hz

• Period ($T$): $T = \frac{1}{f}$

Step-by-Step Guidance

1. Write down the given frequency: $f = 8.66 \times 10^{13}$ Hz.

2. Recall the formula: $T = \frac{1}{f}$.

3. Set up the calculation: Substitute the frequency into the formula.

Try solving on your own before revealing the answer!

Q3. For which of the following equations relating $x$ and $t$ is the motion of the object simple harmonic motion?

• A) $x = 5 \sin 23t$

• B) $x = 8 \cos 3t$

• C) $x = 4 \tan 2t$

• D) $x = 5 \sin 3t$

• E) $x = 2 \cos(3t - 1)$

Background

Topic: Simple Harmonic Motion (SHM)

This question tests your ability to recognize mathematical forms that describe SHM.

Key Terms and Formulas

• SHM is described by equations of the form $x(t) = A \sin(\omega t + \phi)$ or $x(t) = A \cos(\omega t + \phi)$.

• $A$ = amplitude, $\omega$ = angular frequency, $\phi$ = phase constant.

Step-by-Step Guidance

1. Review each equation and check if it matches the SHM form ($\sin$ or $\cos$ with linear argument in $t$).

2. Identify which equations use $\tan$ or other functions (these are not SHM).

3. Check for phase shifts or coefficients in the argument; these are allowed in SHM.

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Q4. The position of an object oscillating on an ideal spring is given by $x = (12.3 \text{ cm}) \cos[(1.26 \text{ s}^{-1}) t]$. At time $t = 0.815$ s, (a) how fast is the object moving? (b) what is the magnitude of the acceleration of the object?

Background

Topic: Simple Harmonic Motion – Position, Velocity, and Acceleration

This question tests your ability to find velocity and acceleration from the position function in SHM.

Key Terms and Formulas

• Position: $x(t) = A \cos(\omega t)$

• Velocity: $v(t) = -A \omega \sin(\omega t)$

• Acceleration: $a(t) = -A \omega^2 \cos(\omega t)$

Step-by-Step Guidance

1. Identify amplitude $A = 12.3$ cm and angular frequency $\omega = 1.26$ s$^{-1}$.

2. Convert amplitude to meters if needed: $A = 0.123$ m.

3. For velocity at $t = 0.815$ s, use $v(t) = -A \omega \sin(\omega t)$ and substitute values.

4. For acceleration at $t = 0.815$ s, use $a(t) = -A \omega^2 \cos(\omega t)$ and substitute values.

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Q5. The x component of the velocity of an object vibrating along the x-axis obeys $v_x(t) = (0.445 \text{ m/s}) \sin[(25.4 \text{ rad/s}) t + 0.223]$. (a) What is the amplitude of the motion of this object? (b) What is the maximum acceleration of the vibrating object?

Background

Topic: SHM – Velocity and Acceleration

This question tests your ability to relate velocity amplitude to displacement amplitude and find maximum acceleration.

Key Terms and Formulas

• Velocity in SHM: $v_x(t) = A \omega \sin(\omega t + \phi)$

• Maximum velocity: $v_{max} = A \omega$

• Maximum acceleration: $a_{max} = A \omega^2$

Step-by-Step Guidance

1. Identify velocity amplitude: $v_{max} = 0.445$ m/s, angular frequency $\omega = 25.4$ rad/s.

2. Use $v_{max} = A \omega$ to solve for amplitude $A$.

3. Use $a_{max} = A \omega^2$ to set up the calculation for maximum acceleration.

Try solving on your own before revealing the answer!

Q6. A 12.0-N object is oscillating in simple harmonic motion at the end of an ideal vertical spring. Its vertical position $y$ as a function of time $t$ is $y(t) = 4.50 \text{ cm} \cos[(19.5 \text{ s}^{-1}) t - \pi/8]$. (a) What is the spring constant of the spring? (b) What is the maximum acceleration of the object? (c) What is the maximum speed that the object reaches? (d) How long does it take the object to go from its highest point to its lowest point?

Background

Topic: Mass-Spring System in SHM

This question tests your ability to use SHM equations to find spring constant, maximum acceleration, maximum speed, and time between extreme positions.

Key Terms and Formulas

• Angular frequency: $\omega = 19.5$ s$^{-1}$

• Amplitude: $A = 4.50$ cm = $0.045$ m

• Weight: $W = 12.0$ N, so $m = \frac{W}{g}$

• Spring constant: $k = m \omega^2$

• Maximum acceleration: $a_{max} = A \omega^2$

• Maximum speed: $v_{max} = A \omega$

• Time from highest to lowest point: $\frac{T}{2}$, where $T = \frac{2\pi}{\omega}$

Step-by-Step Guidance

1. Calculate mass: $m = \frac{12.0}{9.8}$ kg.

2. Use $k = m \omega^2$ to set up the calculation for spring constant.

3. Use $a_{max} = A \omega^2$ for maximum acceleration.

4. Use $v_{max} = A \omega$ for maximum speed.

5. Calculate period $T = \frac{2\pi}{\omega}$, then find time from highest to lowest point: $\frac{T}{2}$.

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Q7. An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period of 0.640 s. What is the spring constant of the spring?

Background

Topic: Mass-Spring System – Period and Spring Constant

This question tests your ability to relate period, mass, and spring constant in SHM.

Key Terms and Formulas

• Period: $T = 0.640$ s

• Weight: $W = 2.450$ N, so $m = \frac{W}{g}$

• Period formula: $T = 2\pi \sqrt{\frac{m}{k}}$

• Rearranged: $k = \frac{4\pi^2 m}{T^2}$

Step-by-Step Guidance

1. Calculate mass: $m = \frac{2.450}{9.8}$ kg.

2. Use the period formula and rearrange to solve for $k$.

3. Set up the calculation: Substitute $m$ and $T$ into $k = \frac{4\pi^2 m}{T^2}$.

Try solving on your own before revealing the answer!

Q8. A 2.25-kg object is attached to a horizontal ideal massless spring on a frictionless table. What should be the spring constant of this spring so that the maximum acceleration of the object will be $g$ when it oscillates with amplitude of 4.50 cm?

Background

Topic: Mass-Spring System – Maximum Acceleration

This question tests your ability to relate maximum acceleration, amplitude, and spring constant.

Key Terms and Formulas

• Maximum acceleration: $a_{max} = A \omega^2$

• Angular frequency: $\omega = \sqrt{\frac{k}{m}}$

• Set $a_{max} = g$

Step-by-Step Guidance

1. Write $a_{max} = A \omega^2 = g$.

2. Substitute $\omega^2 = \frac{k}{m}$ into the equation.

3. Set up the equation: $A \frac{k}{m} = g$ and solve for $k$.

4. Substitute $A = 0.045$ m, $m = 2.25$ kg, $g = 9.8$ m/s$^2$.

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Q9. A 1.6-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 190 N/m. The block is pulled from its equilibrium position at $x = 0.00$ m to a displacement $x = +0.080$ m and is released from rest. The block then executes simple harmonic motion along the horizontal x-axis. What is the velocity of the block at time $t = 0.40$ s?

Background

Topic: SHM – Position and Velocity as Functions of Time

This question tests your ability to use SHM equations to find velocity at a given time.

Key Terms and Formulas

• Position: $x(t) = A \cos(\omega t)$

• Velocity: $v(t) = -A \omega \sin(\omega t)$

• Angular frequency: $\omega = \sqrt{\frac{k}{m}}$

Step-by-Step Guidance

1. Calculate angular frequency: $\omega = \sqrt{\frac{190}{1.6}}$ rad/s.

2. Amplitude $A = 0.080$ m (since released from rest at maximum displacement).

3. Set up velocity equation: $v(t) = -A \omega \sin(\omega t)$.

4. Substitute $t = 0.40$ s into the equation.

Try solving on your own before revealing the answer!

Q10. A building in San Francisco has light fixtures consisting of small 2.35 kg bulbs with shades hanging from the ceiling at the end of light, thin cords 1.50 m long. If a minor earthquake occurs, how many swings per second will these fixtures make?

Background

Topic: Simple Pendulum – Frequency

This question tests your ability to find the frequency of a simple pendulum.

Key Terms and Formulas

• Period: $T = 2\pi \sqrt{\frac{L}{g}}$

• Frequency: $f = \frac{1}{T}$

• Length $L = 1.50$ m, $g = 9.8$ m/s$^2$

Step-by-Step Guidance

1. Set up the period formula: $T = 2\pi \sqrt{\frac{1.50}{9.8}}$.

2. Calculate $T$ and then use $f = \frac{1}{T}$ to find swings per second.

Try solving on your own before revealing the answer!

Q11. A frictionless simple pendulum on Earth has a period of 1.75 s. On Planet X its period is 2.14 s. What is the acceleration due to gravity on Planet X?

Background

Topic: Simple Pendulum – Gravity and Period

This question tests your ability to relate period and gravity for a pendulum.

Key Terms and Formulas

• Period: $T = 2\pi \sqrt{\frac{L}{g}}$

• Compare periods on Earth and Planet X to solve for $g_{X}$.

Step-by-Step Guidance

1. Write the period formula for both planets: $T_{Earth} = 2\pi \sqrt{\frac{L}{g_{Earth}}}$, $T_{X} = 2\pi \sqrt{\frac{L}{g_{X}}}$.

2. Set up the ratio: $\frac{T_{X}}{T_{Earth}} = \sqrt{\frac{g_{Earth}}{g_{X}}}$.

3. Square both sides and solve for $g_{X}$.

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Q12. The angle that a swinging simple pendulum makes with the vertical obeys $\Theta(t) = (0.150 \text{ rad}) \cos[(2.85 \text{ rad/s}) t + 1.66]$. What is the length of the pendulum?

Background

Topic: Simple Pendulum – Angular Frequency and Length

This question tests your ability to relate angular frequency to pendulum length.

Key Terms and Formulas

• Angular frequency: $\omega = \sqrt{\frac{g}{L}}$

• Given $\omega = 2.85$ rad/s

• Rearrange to solve for $L$: $L = \frac{g}{\omega^2}$

Step-by-Step Guidance

1. Write $\omega = 2.85$ rad/s, $g = 9.8$ m/s$^2$.

2. Set up the equation: $L = \frac{g}{\omega^2}$.

3. Substitute values to set up the calculation.

Try solving on your own before revealing the answer!

Q13. A 0.25 kg ideal harmonic oscillator has a total mechanical energy of 4.00 J. If the oscillation amplitude is 20.0 cm, what is the oscillation frequency?

Background

Topic: Energy in SHM – Frequency and Amplitude

This question tests your ability to relate energy, amplitude, mass, and frequency in SHM.

Key Terms and Formulas

• Total energy: $E = \frac{1}{2} m \omega^2 A^2$

• Frequency: $f = \frac{\omega}{2\pi}$

• Amplitude $A = 0.20$ m

Step-by-Step Guidance

1. Write $E = 4.00$ J, $m = 0.25$ kg, $A = 0.20$ m.

2. Set up the energy equation: $E = \frac{1}{2} m \omega^2 A^2$ and solve for $\omega$.

3. Use $f = \frac{\omega}{2\pi}$ to set up the calculation for frequency.

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Q14. A 0.50-kg object is attached to an ideal massless spring of spring constant 20 N/m along a horizontal, frictionless surface. The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position. (a) What is the amplitude of vibration? (b) At what location are the kinetic energy and the potential energy of the system the same?

Background

Topic: Energy in SHM – Amplitude and Energy Distribution

This question tests your ability to relate maximum speed to amplitude and to find the location where kinetic and potential energies are equal.

Key Terms and Formulas

• Maximum speed: $v_{max} = A \omega$

• Angular frequency: $\omega = \sqrt{\frac{k}{m}}$

• At $x = \pm \frac{A}{\sqrt{2}}$, $KE = PE$

Step-by-Step Guidance

1. Calculate $\omega = \sqrt{\frac{20}{0.50}}$ rad/s.

2. Use $v_{max} = A \omega$ to solve for amplitude $A$.

3. Recall that $KE = PE$ when $x = \pm \frac{A}{\sqrt{2}}$.

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Q15. A 1.5-kg mass attached to an ideal massless spring with a spring constant of 20.0 N/m oscillates on a horizontal, frictionless track. At time $t = 0.00$ s, the mass is released from rest at $x = 10.0$ cm. (a) Find the frequency of the oscillations. (b) Determine the maximum speed of the mass. At what point in the motion does the maximum speed occur? (c) What is the maximum acceleration of the mass? At what point in the motion does the maximum acceleration occur? (d) Determine the total energy of the oscillating system. (e) Express the displacement $x$ as a function of time $t$.

Background

Topic: SHM – Frequency, Speed, Acceleration, Energy, and Equation of Motion

This question tests your ability to apply SHM formulas to a mass-spring system.

Key Terms and Formulas

• Frequency: $f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$

• Maximum speed: $v_{max} = A \omega$

• Maximum acceleration: $a_{max} = A \omega^2$

• Total energy: $E = \frac{1}{2} k A^2$

• Displacement: $x(t) = A \cos(\omega t)$ (since released from rest at maximum displacement)

Step-by-Step Guidance

1. Calculate $\omega = \sqrt{\frac{20.0}{1.5}}$ rad/s and $f = \frac{\omega}{2\pi}$.

2. Maximum speed: $v_{max} = A \omega$, with $A = 0.10$ m.

3. Maximum acceleration: $a_{max} = A \omega^2$.

4. Total energy: $E = \frac{1}{2} k A^2$.

5. Write $x(t) = A \cos(\omega t)$.

Try solving on your own before revealing the answer!

Q16. An object of mass 6.8 kg is attached to an ideal massless spring of spring constant 1690 N/m. Calculate the maximum speed the object reaches during its motion.

Background

Topic: SHM – Maximum Speed

This question tests your ability to find maximum speed in SHM given mass and spring constant.

Key Terms and Formulas

• Maximum speed: $v_{max} = A \omega$

• Angular frequency: $\omega = \sqrt{\frac{k}{m}}$

• Amplitude $A$ is needed (not given in question, so may be assumed or provided elsewhere).

Step-by-Step Guidance

1. Calculate $\omega = \sqrt{\frac{1690}{6.8}}$ rad/s.

2. Maximum speed formula: $v_{max} = A \omega$ (you need amplitude $A$).

3. Set up the calculation for $v_{max}$ once $A$ is known.

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Q17. An object weighing 44.1 N hangs from a vertical massless ideal spring. When set in vertical motion, the object obeys $y(t) = (6.20 \text{ cm}) \cos[(2.74 \text{ rad/s}) t - 1.40]$. (a) Find the time for this object to vibrate one complete cycle. (b) What are the maximum speed and maximum acceleration of the object? (c) What is the TOTAL distance the object moves through in one cycle. (d) Find the maximum kinetic energy of the object. (e) What is the spring constant of the spring.

Background

Topic: SHM – Period, Speed, Acceleration, Energy, and Spring Constant

This question tests your ability to use SHM equations for a vertical spring-mass system.

Key Terms and Formulas

• Period: $T = \frac{2\pi}{\omega}$

• Maximum speed: $v_{max} = A \omega$

• Maximum acceleration: $a_{max} = A \omega^2$

• Total distance in one cycle: $4A$

• Maximum kinetic energy: $E_{max} = \frac{1}{2} m v_{max}^2$

• Spring constant: $k = m \omega^2$

• Mass: $m = \frac{44.1}{9.8}$ kg

Step-by-Step Guidance

1. Calculate mass: $m = \frac{44.1}{9.8}$ kg.

2. Period: $T = \frac{2\pi}{2.74}$ s.

3. Maximum speed: $v_{max} = 0.0620 \times 2.74$ m/s (convert amplitude to meters).

4. Maximum acceleration: $a_{max} = 0.0620 \times (2.74)^2$ m/s$^2$.

5. Total distance: $4 \times 0.0620$ m.

6. Maximum kinetic energy: $E_{max} = \frac{1}{2} m v_{max}^2$.

7. Spring constant: $k = m \omega^2$.

Try solving on your own before revealing the answer!