Question

Transverse wave of amplitude $10\text{cm}$ is generated at one end $(x=0)$ of a long string by a tuning fork of frequency $500\text{Hz}$. At a certain instant of time, the displacement of a particle $A$ at $x=100\text{cm}$ is $-5\text{cm}$ and of particle $B$ at $x=200\text{cm}$ is $+5\text{cm}$. What is the wavelength of the wave ?

• A. $2\text{m}$
• B. $3\text{m}$
• C. $4\text{m}$
• D. $5\text{m}$

Answer 1

The correct option is A $2\text{m}$

General equation of a progressive wave travelling in $+x$ direction is
$y=a\sin \left\{\frac{2\pi }{\lambda }\right(vt-x\left)\right\}$
Here, $a=10\text{cm}$
and $\frac{v}{\lambda }=f=500\text{Hz}$

We are given, at time $t$,
$-5=10\sin \left\{\frac{2\pi }{\lambda }\right(vt-100\left)\right\}$
$\Rightarrow \frac{2\pi }{\lambda }(vt-100)=\frac{7\pi }{6}$
and $+5=10\sin \left\{\frac{2\pi }{\lambda }\right(vt-200\left)\right\}$
$\Rightarrow \frac{2\pi }{\lambda }(vt-200)=\frac{\pi }{6}$

Solving these equations give
$\frac{2\pi }{\lambda }\times 100=\pi$
or $\lambda =200\text{cm}=2\text{m}$ which is the choice (a).