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A transverse wave along a string is given by y=2sin(2π(3t−x)+π4), where, x and y are in cm and t in second. The acceleration of a particle located at x = 4cm at t = 1s is:

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a

362π2cm/s2

b

36π2cm/s2

c

762π2cm/s2

d

−36π2cm/s2

answer is D.

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Detailed Solution

We know thatv=dydt =ddt2sin[2π(3t−x)+π4] a=dvdt=ddt12πcos(2π(3t−x)+π4) a−72π2sin(2π(3t−x)+π4) At t =1s and x = 4 cm∴ a=−362π2cm/s2

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A transverse wave along a string is given by y=2sin(2π(3t−x)+π4), where, x and y are in cm and t in second. The acceleration of a particle located at x = 4cm at t = 1s is: