JEE Class main Answered
solve the following
Asked by sarveshvibrantacademy | 02 Apr, 2019, 02:43: PM
Expert Answer
Displacement Y at x and at a time t is given by Y = 2a sin(kx) cos(ωt) ..............................(1)
Given equation for displacement Y = 2 sin(πx) sin( 100πt ) = 2 sin(πx) cos[100πt-(π/2)] ......................(2)
(A) by comparing (1) and (2), we have maximum displacement , 2a sin(kx) = 2 sin(πx)
maximum displacement at x= 1/6 cm , 2sin(π/6) = 1 mm
(B) velocity of the particle is obtained by differentiating eqn.(2),
dY/dt = -2 sin(πx) 100π sin[100πt-(π/2)]
velocity at x = π/6 cm and at t= 1/600 s is = -2 (1/2) 100π sin[ (π/6) - (π/2) ] = 100π (√3/2) = 157√3 mm/s
(C) By comparing eqn.(1) and eqn.(2), we get k = 2π/λ = π or λ = 2 cm
if L is length , for standing wave we have L = nλ/2 , where n = 1,2,3.....
if L = 10 cm and λ = 2 cm , we get n = number of loops = 10
Answered by Thiyagarajan K | 02 Apr, 2019, 03:33: PM
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