Question

The maximum transverse velocity and maximum transverse acceleration of a harmonic wave in a one-dimensional string are 1 $\text{ms}^{-1}$ and 1 $\text{ms}^{-2}$ respectively. The phase velocity of the wave is 1 $\text{ms}^{-1}$. The waveform is

• A. $\sin (x - t)$
• B. $\sin (2x - t)$
• C. $\sin (x - 2t)$
• D. $\sin (x/2 - t)$
• E. $\sin (x - t/2)$

Hint:

The ratio of maximum acceleration to maximum velocity of a particle in a wave always gives the angular frequency $\omega$.

Answer 1

The Correct Option is A

Concept:
For a wave $y = A \sin(kx - \omega t)$: [itemsep=4pt]
• Transverse Velocity ($v_p$): $v_p = \frac{\partial y}{\partial t} = -A\omega \cos(kx - \omega t) \implies |v_{max}| = A\omega$
• Transverse Acceleration ($a_p$): $a_p = \frac{\partial^2 y}{\partial t^2} = -A\omega^2 \sin(kx - \omega t) \implies |a_{max}| = A\omega^2$
• Phase Velocity ($v$): $v = \frac{\omega}{k}$

Step 1: Find $\omega$ and $A$.
Given $|v_{max}| = 1$ and $|a_{max}| = 1$. \[ \frac{A\omega^2}{A\omega} = \frac{1}{1} \implies \omega = 1 \text{ rad/s} \] Then $A(1) = 1 \implies A = 1$ m.

Step 2: Find $k$.
Given phase velocity $v = 1$ ms$^{-1}$. \[ v = \frac{\omega}{k} \implies 1 = \frac{1}{k} \implies k = 1 \text{ m}^{-1} \]

Step 3: Construct the equation.
$y = A \sin(kx - \omega t) = 1 \sin(1x - 1t) = \sin(x - t)$.