Question

A transverse wave travelling on a taut string is represented by :
$y=0.01\sin 2\pi (10t-x)\to y$ and $x$ are in meters and $t$ in seconds. Then

• A. Closest points on the string which differ in phase by ${60}^{\circ }$ are $\left(1/6\right)\text{m}$ apart.
  
• B. The speed of the wave is $10\text{m/s}$
• C. The phase of a certain point on the string changes by ${120}^{\circ }$ in $\left(1/20\right)$ seconds.
  
• D. Maximum particles velocity is $\pi /4\text{m/s}.$

Answer 1

Answer: (A), (B)

The speed of the wave is $10\text{m/s}$

$y=0.01\sin \left[2\pi \right(10t-x\left)\right]$

$\Rightarrow y=0.01\sin [2\pi t-2\pi x]$

Comparing with standard equation, we get
$\omega =20\pi ,k=2\pi$

Wave velocity:

$v=\frac{\omega }{k}=\frac{20\pi }{2\pi }=10\text{m/s}$

$\Delta \varphi =k\Delta x,\text{given}\Delta \varphi ={60}^{\circ }=\frac{\pi }{5}\text{m/s}$

$\frac{\pi }{3}=2\pi \Delta x$

$\Rightarrow \Delta x=\frac{1}{6}\text{m}$

$(v_p{)}_{\text{-max}}=\omega A=20\pi \times 0.01=\frac{\pi }{5}\text{m/s}$

$\Delta {\varphi }_t=\omega \Delta t=20\pi \times \frac{1}{20}=\pi$

so in time $\frac{1}{20}s$, phase changes by and not ${120}^{\circ }$

Hence, $\left(A\right)$ and $\left(B\right)$ are the correct options.