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A body of mass m is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass m is slightly pulled down and released, it oscillates with a time period of 3s. When the mass m is increased by 1kg, the time period of oscillations becomes 5s. The value of m in kg is(A) $\dfrac{9}{{16}}$(B) $\dfrac{3}{4}$(C) $\dfrac{4}{3}$ (D) $\dfrac{{16}}{9}$

Last updated date: 09th Sep 2024
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Hint: This is a question of simple harmonic motion in which the restoring force acting on the body is proportional directly to the displacement of the body. So, we can use the time period expression of simple harmonic motion and calculate the mass.
Formula used:
$T = 2\pi \sqrt {\dfrac{m}{k}}$

$T = 2\pi \sqrt {\dfrac{m}{k}}$ here we can see that T is proportional to root of m
Or, we can say $\dfrac{{{T_1}}}{{{T_2}}} = \sqrt {\dfrac{{{m_1}}}{{{m_2}}}}$
Here ${T_1} = 3\sec $${T_2} = 5\sec$${m_1} = m$ and ${m_2} = m + 1$
$\dfrac{3}{5} = \sqrt {\dfrac{m}{{m + 1}}} \Rightarrow m = \dfrac{9}{{16}}kg$this is the value of m