Question

One end taut string of length $3m$ along the x-axis is fixed at $x=0$. the speed of the waves in the string is $100m{s}^{-1}$. The other end of the string is vibrating in the $y$ direction so that stationary waves are set up in the string. The possible waveform (s) of these stationary waves is (are)

• A. $y\left(t\right)=Asin\frac{2\pi x}{6}cos\frac{50\pi t}{3}$
• B. $y\left(t\right)=Asin\frac{\pi x}{6}cos\frac{100\pi t}{3}$
• C. $y\left(t\right)=Asin\frac{5\pi x}{6}cos\frac{255\pi t}{3}$
• D. $y\left(t\right)=Asin\frac{5\pi x}{6}cos250\pi t$

Answer 1

Answer: (D)

$y\left(t\right)=Asin\frac{5\pi x}{6}cos250\pi t$

The fixed end is a node while the free end is an antinode.

Therefore, at $x=0$ is a node and at $x=3m$ is an antinode. Possible modes of vibration are

$L=(2n+1)\frac{\lambda }{4}$ where $n=0,1,2,3,........$

or $\lambda =\frac{4L}{2n+1}=\frac{12}{2n+1}$ ($\because L=3m$(Given))

$k=\frac{2\pi }{\lambda }=\frac{2\pi }{12/(2n+1)}=\frac{(2n+1)\pi }{6}$

$\omega =vk=100(2n+1)\frac{\pi }{6}=\frac{(2n+1)50\pi }{3}$

For $n=0,k=\frac{\pi }{6}$ $\omega =\frac{50\pi }{3}$

$n=1,k=\frac{5\pi }{2},\omega =50\pi$

$n=2,k=\frac{5\pi }{6},\omega =\frac{250\pi }{3}$

. . .

. . .

. . .

. . .

$n=7,k=\frac{5\pi }{2},\omega =250\pi$

. . .

. . .

. . .

. . .

so on

For $n=0$

$y\left(t\right)=Asin\frac{\pi x}{6}cos\frac{50\pi t}{3}$

For $n=2$

$y\left(t\right)=Asin\frac{5\pi x}{6}cos\frac{250\pi t}{3}$

For $n=7$

$y\left(t\right)=Asin\frac{5\pi x}{2}cos250\pi t$