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Question

A string of length l along x axis is fixed at both ends and is vibrating in second harmonic. If amplitude of incident wave is 2.5 mm, the equation of standing wave is (T is tension and μ is linear density)

A
(2.5mm) sin (2πlx)cos (2π(Tμl2)t)
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B
(5mm) sin (πlx)cos 2πt
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C
(5mm) sin (2πlx)cos (2π(Tμl2)t)
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D
(7.5mm) cos (2πlx)cos (2π(Tμl2)t)
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Solution

The correct option is C (5mm) sin (2πlx)cos (2π(Tμl2)t)
Equation of standing wave fixed at both ends is given as,
y=2a sin(kx) cos (ωt)
Putting a=2.5mm (amplitude of incident wave)

For second harmonic , n=2
l=n×(λ2)
l=2×λ2=λ
Using k=2πλ
k=2πl

Angular frequency
ω=vk=2πlTμ
Substituting value of λ, we get
ω=2πTμl2
Hence equation of standing wave will be,
y=2a sin (2πxl)cos[2πTμl2t]
y=5mm sin(2πlx)cos(2πTμl2t)

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