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One end of a string of length l is tied to the ceiling ofa lift accelerating upwards with an acceleration g/2. The linear mass density of the string is mu(x)=mu_0x^(1//2)where, x is measured from the bottom. The time taken by a pulse to reach from bottom to top is |
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Answer» `sqrt((3l)/g)` WEIGHT of this element `=[mu(x)dx](g+g/2)` Tension at this element , `T_x=int_0^x 3/2 mu(x) gdx` `rArr T_x=3/2 g int_0^x mu_0 x^(1//2) dx` `=(3gmu_0)/2x^(3//2)/(3//2) = g mu_0x^(3//2)` VELOCITY of TRANSVERSE wave at this element `v=sqrt(T/mu) rArr v=(dx)/(dt) = sqrt((gmu_0x^(3//2))/(mu_0x^(1//2)))=sqrt(gx) rArr dt= (dx)/sqrt(gx)` INTEGRATING both sides , we get `int_0^t dt = int_0^l (dx)/sqrt(gx)rArr t= 1/sqrtg int_0^l x^(-1//2) dx rArr t=2 sqrt(l/g)` 
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