Question

A transverse mechanical harmonic wave is travelling on a string. Maximum velocity and maximum acceleration of a particle on the string are $3m/s$ and $90m/s^2$ respectively. If the wave is travelling with a speed of $20m/s$ on the string, then the wave equation is

• A. $y=\sin (30t\pm 1.5x)$
• B. $y=1.5\sin (30t\pm 0.1x)$
• C. $y=0.5\sin (30t\pm 0.1x)$
• D. $y=0.1sin(30t\pm 1.5x)$

Answer 1

The correct option is D $y=0.1sin(30t\pm 1.5x)$

Given:

Maximum velocity of particle $=3m/s$
Maximum acceleration of particle
$=90m/s^2$
Velocity of wave $\left(v\right)=20m/s$
As we know,
Maximum particle velocity,
$v_{max}=\omega a\dots \left(i\right)$
Maximum particle acceleration,
$a_{max}^{\prime }={\omega }^{2}a\dots \left(ii\right)$
From equation $\left(i\right)$ and $\left(ii\right)$
$\frac{a_{max}^{\prime }}{u_{max}^{\prime }}=\omega$
$\Rightarrow \frac{90}{3}=\omega$
$\omega =30rad/s$
$u_{max}=a\omega \Rightarrow 3=a\times 30$
$a=0.1m$
As we know, general equation of wave is
$y=a\sin (\omega t\pm kx)$
As we know,
$k=\frac{\omega }{v}=\frac{30}{20}=1.5m^{-1}$
Hence, by putting the values in equation we get,
$y=0.1\sin (30t\pm 1.5x)$

Final answer: $\left(d\right)$