For an harmonic wave, y = (2.0 cm) sin(3t – 4x) where y and x are in cm and t is in seconds. Match the entries of column I with the entries of column II.
Column-I | Column-II |
(i) The particle at x = 0 at t = 0 | [A] is moving downwards |
(ii) The particle at x = cm at t = 0 | [B] is moving upwards |
(iii) The particle at x = 0 at t = sec. | [C] is accelerated up wards |
(iv) The particle at x = cm at t = sec. | [D] is accelerated downwards |
Text Solution
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Ans.
(i) [B]
(ii) [C]
(iii) [A]
(iv) [D]
Sol.
At t = 0
The shape of the string would be as shown in figure v p at x = 0, would be upwards from,
v p = – v x = 
Acceleration of a particle is given by:
a = – ω 2 y
At x = 0, y = 0, so a = 0
At, x = 
, 
= 0
So, v p = 0, and a = ω 2 A
i.e., accelerated upwards
At t = 
s
y = 2 sin (π– 4x)
= 2 sin 4x
v p|x = 0 = – ve, ie., moving downwards
a p|x = 0 = 0
and 
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