Question

A string fixed at one end only is vibrating in its third harmonic. The wave function is $y(x,t)=0.02sin\left(3.13x\right)cos\left(512t\right)$, where y and x are in metres and t is in seconds. The nodes are formed at positions

• A. (0 m, 2 m)
• B. (0.5 m, 1.5 m)
• C. (0 m, 1.5 m)
• D. (0.5 m, 2 m)

Answer 1

Answer: (B)

(0.5 m, 1.5 m)

Node in formed only at the finest end of the string and the free end acts as an anti node.

In the third harmonics two nodes are formed:

$y(x,t)=0.02\sin \left(3.13x\right)\cos \left(512t\right)$

Standard equation given by

$y(x,t)=2a\sin \left(\frac{2\pi }{\lambda }x\right)\cos \left(2\pi vt\right)$

Comparing both equation, we get

$3.13x=\frac{2\pi }{\lambda }xor,\lambda =\frac{2\lambda }{3.13}=2m\left(approx\right)$

The nodes are formed at $\frac{\lambda }{4}=0.5$ from origin and at $\frac{3\lambda }{4}=1.5$ from origin.