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Question 59 of 66

Q.Two bodies A and B whose masses are in the ratio 1 : 2 are suspended from two separate massless springs of force constants kA and kB respectively. If the two bodies oscillate vertically such that their maximum velocities are in the ratio 1 : 2, the ratio of the amplitude A to that of B is ____.

(a) sqrt(2 kB / kA)
(b) sqrt(kB / 2 kA)
(c) sqrt(8 kB / kA)
(d) sqrt(kB / 8 kA)
Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2024MCQ· 1mImportance★★★★★
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Using v_max = A√(k/m) for each body and the given mass ratio (1:2) and velocity ratio (1:2), the amplitude ratio works out to sqrt(kB/8kA).

For a mass m attached to a spring of constant k, undergoing SHM with amplitude A, the maximum velocity is:

v_max = A ω = A √(k/m)

Let mass of A be m, so mass of B = 2m (given mA : mB = 1 : 2).

Given v_maxA : v_maxB = 1 : 2, i.e. v_maxA / v_maxB = 1/2.

Write:

v_maxA = AA √(kA/m)

v_maxB = AB √(kB/2m)

So: …

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