Physics Exam Review: Oscillations, Waves, Sound, and Light

Q1. When the displacement of a mass on a spring is half the amplitude, what fraction of the mechanical energy is kinetic energy?

Background

Topic: Energy in Simple Harmonic Motion

This question tests your understanding of how mechanical energy is distributed between kinetic and potential energy for a mass-spring system in simple harmonic motion.

Key Terms and Formulas

• Amplitude ($A$): Maximum displacement from equilibrium.

• Displacement ($x$): Position of the mass relative to equilibrium.

• Total mechanical energy ($E$): $E = \frac{1}{2}kA^2$

• Kinetic energy ($K$): $K = E - U$

• Potential energy ($U$): $U = \frac{1}{2}kx^2$

Step-by-Step Guidance

1. Recall that the total mechanical energy in a mass-spring system is constant and given by $E = \frac{1}{2}kA^2$.

2. At displacement $x = \frac{1}{2}A$, calculate the potential energy: $U = \frac{1}{2}k(\frac{1}{2}A)^2$.

3. Simplify the expression for $U$ to relate it to the total energy $E$.

4. Find the kinetic energy by subtracting $U$ from $E$: $K = E - U$.

Try solving on your own before revealing the answer!

Final Answer: 0.75

At $x = \frac{1}{2}A$, the kinetic energy is 75% of the total mechanical energy.

This is because the potential energy at this displacement is $\frac{1}{4}$ of the total energy, so the kinetic energy is $1 - \frac{1}{4} = \frac{3}{4}$.

Q2. At what displacement, as a fraction of amplitude $A$, is the mechanical energy half kinetic and half potential?

Background

Topic: Energy Distribution in Simple Harmonic Motion

This question asks you to find the displacement where kinetic and potential energies are equal in a mass-spring system.

Key Terms and Formulas

• Kinetic energy ($K$): $K = E - U$

• Potential energy ($U$): $U = \frac{1}{2}kx^2$

• Total mechanical energy ($E$): $E = \frac{1}{2}kA^2$

Step-by-Step Guidance

1. Set $K = U$ and use the expressions for kinetic and potential energy.

2. Write $K = E - U$ and $U = \frac{1}{2}kx^2$.

3. Set $E - U = U$ and solve for $x$ in terms of $A$.

4. Express $x$ as a fraction of $A$.

Try solving on your own before revealing the answer!

Final Answer: $x = 0.71A$

At this displacement, kinetic and potential energies are equal.

This is found by solving $\frac{1}{2}kx^2 = \frac{1}{2}kA^2 - \frac{1}{2}kx^2$ and expressing $x$ as a fraction of $A$.

Q3. The common field cricket makes its characteristic chirping sound using a vibrating structure in its wings. The motion and sound intensity can be modeled as a damped oscillation. What is the frequency of oscillations?

Background

Topic: Damped Oscillations

This question tests your ability to analyze a damped oscillation and determine its frequency from a graph of intensity versus time.

Key Terms and Formulas

• Damped oscillation: An oscillation where amplitude decreases over time due to energy loss.

• Frequency ($f$): Number of cycles per second.

• Period ($T$): Time for one cycle, $T = \frac{1}{f}$.

Step-by-Step Guidance

1. Examine the graph and identify the time interval for a complete cycle.

2. Count the number of cycles within a given time span to estimate the period.

3. Calculate the frequency using $f = \frac{1}{T}$.

Try solving on your own before revealing the answer!

Final Answer: 5.0 Hz

The frequency is determined by counting cycles and dividing by the time interval.

Q4. Using a dish-shaped mirror, a solar cooker concentrates the sun's energy onto a pot. Given the intensity of sunlight is about 1000 W/m2, how much solar power does the dish capture?

Background

Topic: Power and Intensity

This question tests your understanding of how to calculate the power captured by a surface given the intensity and area.

Key Terms and Formulas

• Intensity ($I$): Power per unit area, $I = \frac{P}{A}$

• Power ($P$): $P = I \times A$

• Area of a circle: $A = \pi r^2$

Step-by-Step Guidance

1. Find the radius of the dish from its diameter.

2. Calculate the area of the dish using $A = \pi r^2$.

3. Multiply the area by the intensity to find the power: $P = I \times A$.

Try solving on your own before revealing the answer!

Final Answer: 1.5 kW

The power is calculated by multiplying the area of the dish by the intensity of sunlight.

Q5. The figure shows a snapshot graph at t = 0 s of two waves approaching each other at 1 m/s. Choose a correct snapshot graph showing the string at t = 2 s.

Background

Topic: Wave Superposition and Motion

This question tests your ability to predict the position of waves after a certain time, given their speed and direction.

Key Terms and Formulas

• Wave speed ($v$): Distance traveled per unit time.

• Superposition principle: The resultant displacement is the sum of individual displacements.

Step-by-Step Guidance

1. Determine how far each wave will travel in 2 seconds: $\text{distance} = v \times t$.

2. Shift each wave's position accordingly, considering their direction.

3. Draw or select the graph that represents the new positions after 2 seconds.

Try solving on your own before revealing the answer!

Final Answer: See image_8 for the correct graph at t = 2 s.

The waves have moved 2 meters in their respective directions.