MHT CET · Physics · Oscillations

The upper end of the spring is fixed and a mass ' \(\mathrm{m}\) ' is attached to its lower end. When mass is slightly pulled down and released, it oscillates with time period 3 second. If mass ' \(\mathrm{m}\) ' is increased by \(1 \mathrm{~kg}\), the time period becomes 5 second. The value of ' \(\mathrm{m}\) ' is (mass of spring is negligible)

A \(\frac{3}{8} \mathrm{~kg}\)
B \(\frac{5}{9} \mathrm{~kg}\)
C \(\frac{8}{13} \mathrm{~kg}\)
D \(\frac{9}{16} \mathrm{~kg}\)

Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{8}{13} \mathrm{~kg}\)

Step-by-step Solution

Detailed explanation

The formula for the time period of a spring mass system is \(T=2 \pi \sqrt{\frac{m}{k}}\)
For mass \(m+1, T^{\prime}=2 \pi \sqrt{\frac{m+1}{k}}\)
Taking the ratio,
\(\begin{aligned}
& \frac{\mathrm{T}}{\mathrm{T}^{\prime}}=\frac{2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}}{2 \pi \sqrt{\frac{\mathrm{m}+1}{\mathrm{k}}}} \\
& \frac{\mathrm{T}}{\mathrm{T}}=\sqrt{\frac{\mathrm{m}}{\mathrm{m}+1}} \\
& \frac{3}{5}=\sqrt{\frac{m}{m+1}} \\
& \frac{9}{25}=\frac{m}{m+1} \\
& \therefore \quad 9 m+9=25 \mathrm{~m} \\
& \therefore 16 \mathrm{~m}=9 \\
& \therefore \quad \mathrm{m}=\frac{9}{16} \mathrm{~kg} \\
&
\end{aligned}\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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