
A sine wave has an amplitude $A$ and wavelength $\lambda $. Let $V$ be the wave velocity and $v$ be the maximum velocity of a particle in the medium. Which of the following options is correct?
A) $V$ cannot be equal to $v$
B) $V = v$, if $A = 12\pi \lambda $
C) $V = v$, if $A = 2\pi \lambda $
D) $V = v$, if $\lambda = \dfrac{A}{\pi }$
E) $V = v$, if $A = \dfrac{\lambda }{{2\pi }}$
Answer
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Hint: When the velocity of the particle in the medium is maximum, that velocity will be equal to the product of the amplitude of the wave and the angular frequency of the wave. The velocity of the wave is expressed in terms of its frequency and wavelength. Equating the two relations should provide us with the necessary value of the amplitude of the wave.
Formula used: The velocity of the wave is given by, $v = f\lambda $ where $f$ is the frequency of the wave and $\lambda $ is the wavelength of the wave.
The angular frequency of a wave is given by, $\omega = 2\pi f$ where $f$ is the frequency of the wave.
Complete step-by-step answer:
Step 1: List the parameters of the given wave.
The given wave is a sine wave whose amplitude is $A$ and wavelength is $\lambda $ as shown in the figure below.
Let $f$ be the frequency and $\omega $ be the angular frequency of the given sine wave.
The velocity of the wave is also mentioned to be $V$ whereas the maximum velocity of the particle in the medium of propagation is mentioned to be $v$ .
Step 2: Express the relation for the velocity of the wave and the maximum velocity of the particle in the medium.
For the given maximum velocity $v$ of the particle in the medium, we have $v = A\omega $ ------- (1)
Now the velocity of the wave can be expressed as $V = f\lambda $ --------- (2)
Let the two velocities be equal i.e., $v = V$ ---------- (3)
Substituting equations (1) and (2) in (3) we get, $A\omega = f\lambda $ but as $\omega = 2\pi f$ we have $A2\pi f = f\lambda $
$ \Rightarrow 2\pi A = \lambda $ or $A = \dfrac{\lambda }{{2\pi }}$ .
Thus $v = V$ only if $A = \dfrac{\lambda }{{2\pi }}$ .
So the correct option is E.
Note: Here to obtain the relation for the amplitude of the wave for which the velocity of the wave and the maximum velocity of the particle in the medium of propagation are equal, we have to assume that the two velocities are equal. Only then can we derive the necessary relation. For a sine wave, the frequency of the wave and its wavelength are inversely proportional to each other.
Formula used: The velocity of the wave is given by, $v = f\lambda $ where $f$ is the frequency of the wave and $\lambda $ is the wavelength of the wave.
The angular frequency of a wave is given by, $\omega = 2\pi f$ where $f$ is the frequency of the wave.
Complete step-by-step answer:
Step 1: List the parameters of the given wave.
The given wave is a sine wave whose amplitude is $A$ and wavelength is $\lambda $ as shown in the figure below.

Let $f$ be the frequency and $\omega $ be the angular frequency of the given sine wave.
The velocity of the wave is also mentioned to be $V$ whereas the maximum velocity of the particle in the medium of propagation is mentioned to be $v$ .
Step 2: Express the relation for the velocity of the wave and the maximum velocity of the particle in the medium.
For the given maximum velocity $v$ of the particle in the medium, we have $v = A\omega $ ------- (1)
Now the velocity of the wave can be expressed as $V = f\lambda $ --------- (2)
Let the two velocities be equal i.e., $v = V$ ---------- (3)
Substituting equations (1) and (2) in (3) we get, $A\omega = f\lambda $ but as $\omega = 2\pi f$ we have $A2\pi f = f\lambda $
$ \Rightarrow 2\pi A = \lambda $ or $A = \dfrac{\lambda }{{2\pi }}$ .
Thus $v = V$ only if $A = \dfrac{\lambda }{{2\pi }}$ .
So the correct option is E.
Note: Here to obtain the relation for the amplitude of the wave for which the velocity of the wave and the maximum velocity of the particle in the medium of propagation are equal, we have to assume that the two velocities are equal. Only then can we derive the necessary relation. For a sine wave, the frequency of the wave and its wavelength are inversely proportional to each other.
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