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Question

Physics Question on Waves

The equation of stationary wave along a stretched string is given by y=5sinπx3cos40πty=5\, \sin \frac{\pi x}{3} \cos 40\, \pi t, where, xx and yy are in cmcm and tt in second. The separation between two adjacent nodes is :

A

1. 5 cm

B

3 cm

C

6 cm

D

4 cm

Answer

3 cm

Explanation

Solution

Given, equation
y=5sinπx3cos40πty=5\, \sin \frac{\pi x}{3} \cos 40 \pi t
Comparing with standard equation
y=Asinkxcosωty=A \sin k x \cos \omega t
k=π3\therefore k =\frac{\pi}{3}
2πλ=π3\Rightarrow \frac{2 \pi}{\lambda} =\frac{\pi}{3}
λ=6cm\lambda =6\, cm
Separation between two consecutive nodes
=λ2=62=3cm=\frac{\lambda}{2}=\frac{6}{2}=3\, cm