Question

A stationary wave set up on a string have the equation $y=\left(2\text{mm}\right)\left[\text{cos}\right(6.28{\text{m}}^{-1}\left)\text{x cos}\right(\omega t\left)\right]$. This stationary wave is created by two identical waves, of amplitude $A$ each moving in opposite directions along the string. Then,

• A. $A=2\text{mm}$
• B. $A=4\text{mm}$
• C. The smallest length of the string is $50\text{cm}$
• D. The smallest length of the string is $25\text{cm}$

Answer 1

The correct option is D The smallest length of the string is $25\text{cm}$

From the equation of stationary wave
$y=2\text{mm}\left[\text{cos}\right(6.28{\text{m}}^{-1}\left)\text{x cos}\right(\omega t\left)\right]$
Comparing it with
$y=2\text{A cos kx cos}\omega t$
$2A=2\text{mm}\Rightarrow A=1\text{mm}$ (Amplitude of parent wave)
$k=6.28=2\pi$

Put $(x=0)$ at $(t=0)$ in above equation, It gives
$y=2\text{mm}$
That represents $x=0$ is antinode

Minimum possible length of string is,
$L=\frac{\lambda }{4}$ (Distance between Node and Antinode)
$\Rightarrow L_{min}=\frac{\lambda }{4}=\left(\frac{\frac{2\pi }{k}}{4}\right)$
$\therefore L_{min}=\frac{1}{4}\text{m}=25\text{cm}$