JEE Advanced | 2013Wave motionHard
Question
A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation,
y(x, t) = (0.01m) sin [(62.8 m-1)x]cos [(628 s-1)t].
Assuming π = 3.14, the correct statement(s) is (are)
y(x, t) = (0.01m) sin [(62.8 m-1)x]cos [(628 s-1)t].
Assuming π = 3.14, the correct statement(s) is (are)
Options
A.The number of nodes is 5.
B.The length of the string is 0.25 m.
C.The maximum displacement of the midpoint of the string, from its equilibrium position is 0.01m.
D.The fundamental frequency is 100 Hz.
Solution
Given equation y = 0.01 sin [20 πx] cos [200 πt]
length = 5
no. of nodes = 4 + 2 = 6
Maximum displacement = 0.01
5th harmonic frequency = 100 Hz
⇒ Fundamental frequency = 20 Hz
length = 5
no. of nodes = 4 + 2 = 6
Maximum displacement = 0.01
5th harmonic frequency = 100 Hz
⇒ Fundamental frequency = 20 Hz
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