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Last updated date: 01st Dec 2023
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# A mass m attached to a spring oscillates every 2 sec. If the mass is increased by 2 kg, then time-period increases by 1 sec. The initial mass isA. 1.6 kgB. 3.9 kgC. 9.6 kgD. 12.6 kg

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Hint: When the mass is attached to the spring, then the motion of the mass is simple harmonic motion with constant time period. The time taken to complete one oscillation is time period. Depending on the spring constant and the mass of the body attached the time period varies.

Formula used:
$T = 2\pi \sqrt {\dfrac{m}{k}}$, here T is the period, k is the spring constant and m is the mass of the block.

Complete step by step solution:
For the initial case;
Let the spring constant of the spring used in the oscillation is K. The value of the mass is given as m. And the period of oscillator is 2sec

Then using the formula of period of spring-mass oscillator is,
$T = 2\pi \sqrt {\dfrac{m}{k}}$
$2\sec = 2\pi \sqrt {\dfrac{m}{k}}$

For the final case;
The mass of the block is increased by 2 kg, then the final mass of the block is $\left( {m + 2} \right)kg$. The spring constant is same as before.

So, using the formula of period of spring-mass oscillator, the final period is 1 sec greater than the initial period.

So, the final time period is ${T_2} = 3\sec$. Using the period formula,
${T_2} = 2\pi \sqrt {\dfrac{{{m_2}}}{{{k_2}}}}$
$3\sec = 2\pi \sqrt {\dfrac{{m + 2}}{k}}$

On dividing the first case period with the second period, we get
$\dfrac{{2\sec }}{{3\sec }} = \dfrac{{\left( {2\pi \sqrt {\dfrac{m}{k}} } \right)}}{{2\pi \sqrt {\dfrac{{m + 2}}{k}} }}$
$\dfrac{2}{3} = \sqrt {\left( {\dfrac{m}{{m + 2}}} \right)}$

On squaring both the sides, we get
$\dfrac{4}{9} = \dfrac{m}{{m + 2}}$
$9m = 4m + 8$
$5m = 8$
$m = \dfrac{8}{5}$
$m = 1.6$
So, the initial mass is $1.6kg$

Therefore, the correct option is (A).

Note: In the question the mass is increased by 2 kg, not twice. We should be careful while writing the relation between the final and initial quantity.