JEE Class main Answered
Plz solve my question given below
Asked by sarveshvibrantacademy | 16 Mar, 2019, 13:47: PM
Expert Answer
(i) Equation for displacement at various point at various time is given by, y = 10-4 sin [ ( πx/10 ) - (50 π t ) ] ...................(1)
when t = 1/50 s, y = 10-4 sin [ ( πx/10 ) - π ] = - 10-4 sin [ π - ( πx/10 ) ] = 10-4 sin [ π x / 10 ] ....................(2)
graph of eqn.(2) is given above.
(ii) Average power , P = (1/2)μω2 A2 λ ............................(3)
where μ is linear mass density of the string ( mass per unit length), ω = 2π v/λ is the angular frequency, v is the velocity of wave in the string,
A is the vibration amplitude and λ is wavelength
P = (1/2) μ ( 4π2 v2 / λ2 ) A2 λ ...........................(4)
in stretched string under tension T, velocity of the wave is given by, 
.
. By substituting for v in eqn.(4) and after simplification we get, P = (1/2) ( 4π2 / λ ) T A2 . Wavelength λ = 20 m as seen from the above graph.
By substituting all the values , we get P = (1/2) ( 4π2 / 20)×100×10-8 W ≈ 10-6 W or nearly 1 micro-Watt
(iii) Velocity - time graph of particle at x = 5m
By differentiating eqn.(1), we get (dy/dt) = v = 0.01 cos [ ( πx/10 ) - (50 π t ) ] × (-50π) = (-π/2) cos [ ( πx/10 ) - (50 π t ) ]
velocity at x = 5 m is given by, vx=5 = (-π/2) cos [ ( π/2 ) - (50 π t ) ] = (-π/2) sin [ 50 π t ]
the graph is shown below
Answered by Thiyagarajan K | 17 Mar, 2019, 20:52: PM
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