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AP Physics 2- 14.2 Periodic Waves - Exam Style questions - FRQs- New Syllabus

Periodic Waves AP  Physics 2 FRQ

Unit 14: Waves , Sound , and Physical Optics 

Weightage : 15–18%

AP Physics 2 Exam Style Questions – All Topics

Question

A transverse wave travels to the right along a string.
(a) Two dots have been painted on the string. In the diagrams below, those dots are labeled \(P\) and \(Q\).
i. The figure below shows the string at an instant in time. At the instant shown, dot \(P\) has maximum displacement and dot \(Q\) has zero displacement from equilibrium. At each of the dots \(P\) and \(Q\), draw an arrow indicating the direction of the instantaneous velocity of that dot. If either dot has zero velocity, write \(v=0\) next to the dot.
ii. The figure below shows the string at the same instant as shown in part (a)(i). At each of the dots \(P\) and \(Q\), draw an arrow indicating the direction of the instantaneous acceleration of that dot. If either dot has zero acceleration, write \(a=0\) next to the dot.
The figure below represents the string at \(t=0\), the same instant as shown in part (a) when dot \(P\) is at its maximum displacement from equilibrium. For simplicity, dot \(Q\) is not shown.
(b)
i. On the grid below, draw the string at a later time \(t=T/4\), where \(T\) is the period of the wave.
Note: Do any scratch work on the grid at the bottom of the page. Only the sketch made on the grid immediately below will be graded.
ii. On your drawing above, draw a dot to indicate the position of dot \(P\) on the string at time \(t=T/4\) and clearly label the dot with the letter \(P\).
(c) Now consider the wave at \(t=T\). Determine the distance traveled, not the displacement, by dot \(P\) between times \(t=0\) and \(t=T\).

Most-appropriate topic codes (AP Physics \(2\)):

• Topic \(14.1\) — Periodic Waves (Part \( \mathrm{(a)} \), Part \( \mathrm{(b)} \), Part \( \mathrm{(c)} \))
• Topic \(14.2\) — Wave Properties (Part \( \mathrm{(a)} \), Part \( \mathrm{(b)} \), Part \( \mathrm{(c)} \))
▶️ Answer/Explanation

(a)(i)
Dot \(P\) is at maximum displacement, so its instantaneous vertical velocity is zero.

\(\boxed{v_P=0}\)

For a wave traveling to the right, the string’s vertical velocity is opposite in sign to the local slope of the wave. At \(Q\), the wave is sloping downward as \(x\) increases, so dot \(Q\) is moving upward.

\(\boxed{\text{At }Q\text{, draw an upward velocity arrow.}}\)

(a)(ii)
For transverse wave motion, each point on the string accelerates toward the equilibrium position. Dot \(P\) is below equilibrium at maximum displacement, so its acceleration is upward.

\(\boxed{\text{At }P\text{, draw an upward acceleration arrow.}}\)

Dot \(Q\) is at the equilibrium position, so its instantaneous acceleration is zero.


\(\boxed{a_Q=0}\)

(b)(i)
In a time \(T/4\), a wave traveling to the right moves horizontally by one-fourth of a wavelength.

From the graph, one wavelength is approximately \(24\,\text{cm}\). Therefore, the wave shifts to the right by

\(\dfrac{\lambda}{4}=\dfrac{24\,\text{cm}}{4}=6\,\text{cm}\)

So the sketch at \(t=T/4\) should have the same shape and amplitude as the original wave, but shifted \(6\,\text{cm}\) to the right.


\(\boxed{\text{Draw the original wave shifted right by }\lambda/4.}\)

(b)(ii)
Dot \(P\) is a point on the string, not a point that travels horizontally with the wave shape. At \(t=0\), dot \(P\) is at a maximum downward displacement. After one-fourth period, that dot reaches equilibrium.

Therefore, dot \(P\) should be placed at the same horizontal location as before, but on the equilibrium line. From the graph, this is at about \(x=18\,\text{cm}\).

\(\boxed{\text{Place }P\text{ at }x\approx18\,\text{cm}\text{ on }y=0.}\)

(c)
The amplitude of the wave is \(8\,\text{cm}\). Dot \(P\) starts at maximum downward displacement, moves to equilibrium, then maximum upward displacement, then back to equilibrium, and finally returns to maximum downward displacement after one full period.

The total distance traveled by dot \(P\) in one full cycle is four amplitudes:

\(d=4A\)

\(d=4\left(8\,\text{cm}\right)\)

\(\boxed{d=32\,\text{cm}}\)

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