Question

A wire is stretched and fixed at two ends. Transverse stationary waves are formed in it. It oscillates in its third overtone mode. The equation of stationary wave is
$Y=Asinkxcos\omega t$
Choose the correct options.

• A. Amplitude of constituent waves is $\frac{A}{2}$
• B. The wire oscillates in three loops
• C. The length of wire is $\frac{4\pi }{k}$
• D. Speed of stationary wave is $\frac{\omega }{k}$

Answer 1

Answer: (A), (C)

The correct options are
A Amplitude of constituent waves is $\frac{A}{2}$
C The length of wire is $\frac{4\pi }{k}$

We see that the stationary wave is obtained by the superposition of
$y_1=\frac{A}{2}sin(kx-\omega t)$
$y_2=\frac{A}{2}sin(kx+\omega t)$
i.e. amplitude of the constituent waves is $\frac{A}{2}$
The wire is in third overtone(fourth harmonic) mode of oscillation (4 loops).
Hence length of wire $=2\lambda$
but $k=\frac{2\pi }{\lambda }$
i.e.
Length of wire $=\frac{4\pi }{k}$
Speed of stationary wave is zero.
Hence options A C are correct.