JEE MAIN - Physics (2021 - 24th February Morning Shift - No. 2)
In the given figure, a mass M is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is k. The mass oscillates on a frictionless surface with time period T and amplitude A. When the mass is in equilibrium position, as shown in the figure, another mass m is gently fixed upon it. The new amplitude of oscillations will be :
_24th_February_Morning_Shift_en_2_1.png)
$$A\sqrt {{M \over {M - m}}} $$
$$A\sqrt {{{M - m} \over M}} $$
$$A\sqrt {{{M + m} \over M}} $$
$$A\sqrt {{M \over {M + m}}} $$
Explanation
Given, initial amplitude = A
Velocity at mean position, v = A$$\omega$$
Applying conservation of momentum at mean position, we get
M1v1 = M2v2
MA$$\omega$$ = (M + m)v'
$$ \Rightarrow v' = {{MA\omega } \over {M + m}} = {{MA\sqrt {{k \over M}} } \over {M + m}}$$
$$\therefore$$ $$v' = A'\omega ' = A'\sqrt {{k \over {M + m}}} $$
$$ \Rightarrow A' = {{MA\sqrt {{k \over M}} } \over {M + m}} \times \sqrt {{{M + m} \over k}} $$
$$A' = \sqrt {{M \over {M + m}}} A$$
Velocity at mean position, v = A$$\omega$$
Applying conservation of momentum at mean position, we get
M1v1 = M2v2
MA$$\omega$$ = (M + m)v'
$$ \Rightarrow v' = {{MA\omega } \over {M + m}} = {{MA\sqrt {{k \over M}} } \over {M + m}}$$
$$\therefore$$ $$v' = A'\omega ' = A'\sqrt {{k \over {M + m}}} $$
$$ \Rightarrow A' = {{MA\sqrt {{k \over M}} } \over {M + m}} \times \sqrt {{{M + m} \over k}} $$
$$A' = \sqrt {{M \over {M + m}}} A$$
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