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=2πl√Tμ
Substituting value of λ, we get ω=2π√Tμl2
Hence equation of standing wave will be, y=2a sin (2πxl)cos[2π√Tμl2t] ∴y=5mmsin(2πlx)cos(2π√Tμl2t)