JEE Advanced | 2013SHMHard
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
Create a free account to view solution
View Solution FreeMore SHM Questions
A ball is hung vertically by a thread of length ′l′ from a point ′P′ of an inclined wall that ma...The displacement of a particle in S.H.M. is x = a sinωt. Which of the following graph between displacement and time...Values of the acceleration of a particle moving in simple harmonic motion as a function of its displacement x are given ...The displacement of a particle varies according to the relation x = 3 sin 100 t + 8 cos250 t. Which of the following is ...On loading a spring with bob, its period of oscillation in a vertical plane is T. If this spring pendulum is tied with o...