A string of length ′l′ is fixed at both ends. It is vibrating in its 3rd overtone with maximum amplitude ′a′. The amplitude at a distance l3 from one end is:
A
a
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B
0
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C
√3a2
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D
a2
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Solution
The correct option is D√3a2 For a string vibrating in its nth overtone (n+1)tj harmonic) y=2Asin((n+1)πxL)cosωt For x=l3.2A=a and n=3; y=[asin(4πl⋅l3)]cosωt =asin4π3cosωt=−a(√32)cosωt i.e. at x=l3, the amplitude is √3a2.