A mass m 1 connected to a horizontal spring performs S.H.M. with amplitude A . While mass m 1 is passing through mean position another mass m 2 is placed on it so that both the masses move together with amplitude A 1 . The ratio of A 1 A is ( m 2 < m 1 )

Step-by-step solution

P.E of the oscillating mass is given by E = 1 2 m ω x 2 = 1 2 k x 2 where k = m ω 2 Let k = m ω 2 be the velocity at mean position. When mass M 2 is attached v 2 = m 1 v 1 m 1 + m 2 If A 1 be the new amplitude 1 2 k A 1 2 = 1 2 ( m 1 + m 2 ) v 2 2 1 2 ( m 1 + m 2 ) m 1 2 v 1 2 ( m 1 + m 2 ) 2 = 1 2   m 1 v 1 2 × m 1 m 1 + m 2 = 1 2   k A 2 × m 1 m 1 + m 2 ⇒ A 1 A = m 1 m 1 + m 2