Wave motionMedium
Question
One end of a string of length L is tied to the ceiling of lift accelerating upwards with an accelerating 2g. The other end of the string is free. The linear mass density of the string varies linearly from 0 of λ from bottom to top : -
Options
A.The velocity of the wave in the string will be 0
B.The acceleration of the wave on the string will be 3g/4 every where.
C.The time taken by a pulse to reach from bottom to top will be 
D.The time taken by a pulse to reach from bottom to top will be 
Solution
μ = 
x (at a distance ′x′ from free end)
∴ T =
μdx (g + 2g) = 
∴ vwave =
⇒ v2 =
⇒ 2v 
⇒ a = 3g / 4
(constant everywhere)
Now S = ut +
at2 ⇒ L = 0 + 
⇒ t =
x (at a distance ′x′ from free end)∴ T =
μdx (g + 2g) = ∴ vwave =
⇒ v2 =
⇒ 2v ⇒ a = 3g / 4(constant everywhere)
Now S = ut +
at2 ⇒ L = 0 + ⇒ t =
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