Question

The equation of a wave is $y = 5\sin \left( {\dfrac{t}{{0 \cdot 04}} - \dfrac{x}{4}} \right)$, where, x is in cm and t is in second. The maximum velocity of the wave will be:
A. $1m{s^{ - 1}}$
B. $2m{s^{ - 1}}$
C. $1 \cdot 5m{s^{ - 1}}$
D. $1 \cdot 25m{s^{ - 1}}$

Answer 1

Hint: For any harmonic wave, the waveform is always a sinusoidal function of space coordinate and time. Coefficient of time is given by angular frequency and wave number is the coefficient of space term. Maximum velocity of the wave is obtained by differentiating the equation of displacement of the wave.
Formulae used:
General function of any waveform is given as:
$y = A\sin (\omega t - kx)$ …...(1)
Where,
y gives the waveform,
A is amplitude of the wave,
k is wave number,
x denotes the position of the particle,
$\omega $is the angular velocity,
t denotes the time.

Complete answer:
The equation for the displacement of particle from its mean position is given by –
$y = A\sin (\omega t - kx)$
If we differentiate this equation with respect to time, we obtain the value of velocity of the wave.
$v = \dfrac{{dy}}{{dt}} = \dfrac{d}{{dt}}\left( {A\sin (\omega t - kx)} \right)$
Differentiating, we get –
$v = A\omega \dfrac{d}{{dt}}\left( {\sin (\omega t - kx)} \right)$
$ \Rightarrow v = A\omega \cos (\omega t - kx)$
The maximum value of the velocity will be at an instant where the function $\cos \left( {\omega t - kx} \right) = 1$
Substituting the value of maximum velocity, we obtain the value of maximum velocity of the wave as –
$V = A\omega $
Given: The equation of the wave is given as $y = 5\sin \left( {\dfrac{t}{{0 \cdot 04}} - \dfrac{x}{4}} \right)$
Comparing the given equation of the wave with the general form of eq.(1) to obtain the value of $\omega $ and amplitude A as:
$\omega = \dfrac{1}{{0 \cdot 04}} = 25{s^{ - 1}}$
$A = 5cm$
Substituting the values in the equation of velocity, we get –
$V = A\omega $
$ \Rightarrow V = A\omega = 5 \times 25 = 125cm{s^{ - 1}}$
$\therefore V = \dfrac{{125}}{{100}}m{s^{ - 1}} = 1 \cdot 25m{s^{ - 1}}$

Hence, the correct option is Option D.

Note:
In this problem there is a big chance for a student to make a mistake. Here, the problem has asked for the wave velocity, not the velocity of the particles. Wave velocity is the velocity the whole wave propagates, where particle velocity is the velocity of each particle of the medium through which the wave is propagating. Particle velocity is obtained by the ratio of the quantities $\omega $ and k in the wave equation w.r.t time and that is not asked here.