Question
Question: A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant tha...
A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that it is at a distance 32A from equilibrium position. The new amplitude of the motion is:
A) 3A41
B) 3A
C) A3
D) 37A
Solution
Recall that the simple harmonic motion is defined as the motion in which the restoring force is directly proportional to the displacement of the body from its mean position. The direction of the restoring force is towards its mean position.
Complete step by step solution:
The velocity of a particle executing SHM at any instant, is the time rate of change of its displacement at that instant.
v=ωA2−x2
Where ωis the angular frequency, A is the amplitude and x is the displacement of the particle.
Suppose that the new amplitude of the motion be ‘A’.
If A is the initial amplitude andωis the angular frequency. Initial velocity of a particle performs SHM,
⇒v2=ω2[A2−(32A)2]---(i)
The final velocity of the particle is given by
⇒(3v)2=ω2[A′2−(32A)2]---(ii)
The ratio of equation (i) and (ii) is written as,
⇒(3v)2v2=A′2−94A2A2−94A2
⇒91=A′2−94A299A2−4A2
⇒91=A′2−94A295A2
⇒91(A′2−94A2)=95A2
\Rightarrow {A' ^2} - \dfrac{{4{A^2}}}{9} = \dfrac{{5{A^2}}}{9} \times 9
\Rightarrow {A' ^2} = 5{A^2} + \dfrac{{4{A^2}}}{9}
\Rightarrow {A' ^2} = \dfrac{{45{A^2} + 4{A^2}}}{9}
⇒A’2=949A2
⇒A′=37A
The new amplitude of the motion is 37A.
Option D is the right answer.
Note: It is important to note that all the simple harmonic motions are periodic but all periodic motions are not simple harmonic. This is because a simple harmonic motion is a type of periodic motion in which there is to and fro movement of an object about its mean position. But the mean position of the object can be different. But periodic motion is a type of motion which repeats itself after fixed intervals of time. For example the motion of Earth around the sun is a type of periodic motion as it repeats after a fixed interval of time.