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Question

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

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

The correct option is C (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)=0.02sin(3.13x)cos(512t)

Standard equation given by

y(x,t)=2asin(2πλx)cos(2πvt)

Comparing both equation, we get

3.13x=2πλxor,λ=2λ3.13=2m(approx)

The nodes are formed at λ4=0.5 from origin and at 3λ4=1.5 from origin.


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