Q11 / 107 · NTA Abhyas (JEE Main) 2020 · Physics · Oscillations

A mass m1=1 kg connected to a horizontal spring performs S.H.M. with amplitude A. While mass m1 is passing through mean position another mass m2=3 kg is placed on it so that both the masses move together with amplitude A1. The ratio of A1A is pq12, where p and q are the smallest integers. Then what is the value of p+q?

Correct answer: 5

Solution

P.E of the oscillating mass is given by

  E= 1 2 mω 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  k A 2 × m 1 m 1 + m 2

⇒ A 1 A = m 1 m 1 + m 2 =14

Hence, the correct answer is 5.

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