Physics with Calculus Exam 1 Practice Guidance: Waves, SHM, and Electric Fields

Q3. The frequency of the wave shown in the figure is:

Background

Topic: Wave Properties (Frequency and Period)

This question tests your ability to determine the frequency of a wave from its graphical representation, specifically by analyzing the period shown on a displacement vs. time graph.

Key Terms and Formulas

• Period (T): The time it takes for one complete cycle of the wave.

• Frequency (f): The number of cycles per second, measured in Hertz (Hz).

• Relationship:

$f = \frac{1}{T}$

Step-by-Step Guidance

1. Examine the graph to determine the period $T$ of the wave. The period is the time interval between two consecutive points where the wave repeats its pattern (e.g., from peak to peak).

2. Identify the value of $T$ by looking at the time axis. For example, if the wave completes one cycle in 2 seconds, then $T = 2$ s.

3. Recall the formula for frequency: $f = \frac{1}{T}$.

4. Set up the calculation for frequency using the period you found from the graph.

Try solving on your own before revealing the answer!

Final Answer: 0.5 Hz

From the graph, the period $T$ is 2 seconds. Using $f = \frac{1}{T}$, $f = \frac{1}{2} = 0.5$ Hz.

This means the wave completes half a cycle per second.

Q4. In the figure, which curve best represents the variation of wave speed $V$ as a function of tension for transverse waves on a stretched string?

Background

Topic: Wave Speed on a Stretched String

This question tests your understanding of how the speed of a transverse wave on a string depends on the tension in the string.

Key Terms and Formulas

• Wave speed ($V$): The speed at which a wave travels along the string.

• Tension ($T$): The force stretching the string.

• Mass per unit length ($\mu$): The mass of the string divided by its length.

• Formula:

$V = \sqrt{\frac{T}{\mu}}$

Step-by-Step Guidance

1. Recall that the speed of a wave on a string is proportional to the square root of the tension: $V \propto \sqrt{T}$.

2. Examine the curves in the figure. Identify which curve shows a square root relationship (i.e., increases rapidly at first, then more slowly as tension increases).

3. Compare the shapes of the curves to the expected mathematical form of $V$ vs. $T$.

4. Determine which curve best matches the theoretical relationship.

Try solving on your own before revealing the answer!

Final Answer: Curve B

Curve B best represents the relationship $V = \sqrt{T/\mu}$, showing the correct square root dependence of wave speed on tension.

This is because the wave speed increases with tension, but not linearly—it follows a square root curve.

Q12. The position of an object oscillating on an ideal spring is given by $x(t) = (0.10\,\text{m})\cos[(1.26\,\text{rad/s})\,t]$. At time $t = 0.8$ s, what is the magnitude of the acceleration of the object?

Background

Topic: Simple Harmonic Motion (SHM) and Acceleration

This question tests your ability to calculate the acceleration of an object in SHM at a specific time, using the position function.

Key Terms and Formulas

• Position function: $x(t) = A \cos(\omega t)$

• Angular frequency ($\omega$): The rate at which the object oscillates, in rad/s.

• Acceleration in SHM: $a(t) = -\omega^2 x(t)$

Step-by-Step Guidance

1. Write the position function: $x(t) = 0.10 \cos(1.26 t)$, where $A = 0.10$ m and $\omega = 1.26$ rad/s.

2. Recall the formula for acceleration in SHM: $a(t) = -\omega^2 x(t)$.

3. Calculate $x(0.8)$ by substituting $t = 0.8$ s into the position function.

4. Set up the calculation for $a(0.8)$ using the value of $x(0.8)$ and $\omega$.

Try solving on your own before revealing the answer!

Final Answer: 0.1 m/s2

Using $a(t) = -\omega^2 x(t)$, substitute $\omega = 1.26$ rad/s and $x(0.8)$ from the position function. The magnitude of the acceleration is $0.1$ m/s2.

This shows how the acceleration depends on both the angular frequency and the instantaneous position.