Question

A mass M is moving in SHM on a line with amplitude A and frequency f in a spring mass system. Suddenly half of the mass comes to rest just at the moment when it crosses mean position then the new amplitude becomes λA, then λ will be?

Answer 1

Answer:

When a body undergoing SHM comes at mean position, the total force acting on it becomes zero. Hence at the mean position, we can conserve momentum. Applying conservation of momentum, we get:

→ mv = 0.5mv'  [Given in the question that mass becomes half]

Here, v is the inital speed before loss of mass, and v' is the final speed after losing half of the mass

From this we get:

→ v = 0.5v'

→ v' = 2v   ... (1)

Now we know that, at the mean position, the velocity of the particle is maximum which has a value of Aw. Hence writing v and v' in terms of Aw we get:

→ A'w' = 2Aw

$\rightarrow (\lambda A).\omega ' = 2A.\omega\\\\\rightarrow \omega = \sqrt{ \dfrac{k}{m}}\\\\\\\rightarrow (\lambda A)\:\sqrt{\dfrac{k}{0.5m}} = 2A.\:\sqrt{\dfrac{k}{m}}\\\\\\\rightarrow \lambda.\sqrt{2} = 2\\\\\\\boxed{ \lambda = \dfrac{2}{\sqrt{2}} = \sqrt{2}}$

Hence the value of λ is √2.