JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Elasticity & SHM

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Two massless springs with spring constants 2 k and 9 k , carry 50 g and 100 g masses at their free ends. These two masses oscillate vertically such that their maximum velocities are equal. Then, the ratio of their respective amplitudes will be :

(A) $1 : 2$
(B) $3 : 2$ ✓ Correct
(C) $3 : 1$
(D) $2 : 3$

Answer: (B)

Solution

In vertical simple harmonic motion the maximum speed is

v_{m a x} = ω A = A k / m

, so equating the two maximum speeds gives

A_{1} \frac{2 k}{50} = A_{2} \frac{9 k}{100}

, using the spring constants

2 k

and

9 k

with the masses

50 g

and

100 g

. Rearranging,

\frac{A _{1}}{A _{2}} = \frac{9 k /100}{2 k /50} = \frac{9 \times 50}{2 \times 100} = \frac{9}{4} = \frac{3}{2}

, so the amplitudes are in the ratio

3 : 2

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Single correct option

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