Question:
A mass \(x\) gram is suspended from a light spring. It is pulled in downward direction and released so that mass performs S.H.M. of period T. If mass is increased by Y gram, the period becomes \(\frac{4 \mathrm{~T}}{3}\). The ratio of \(\mathrm{Y} / \mathrm{x}\) is
A mass \(x\) gram is suspended from a light spring. It is pulled in downward direction and released so that mass performs S.H.M. of period T. If mass is increased by Y gram, the period becomes \(\frac{4 \mathrm{~T}}{3}\). The ratio of \(\mathrm{Y} / \mathrm{x}\) is
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The period of a spring-mass system is proportional to the square root of the mass.
Updated On: Nov 13, 2026
- \(7:9\)
- \(5:4\)
- \(3:2\)
- \(8:7\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept
\(T=2\pi\sqrt{\dfrac{m}{k}}\), so \(T\propto\sqrt{m}\).
Step 2: Set up the ratio
\(\dfrac{T'}{T}=\sqrt{\dfrac{x+Y}{x}}=\dfrac43\).
Step 3: Square
\(\dfrac{x+Y}{x}=\dfrac{16}{9}\Rightarrow1+\dfrac{Y}{x}=\dfrac{16}{9}\).
Step 4: Solve
\(\dfrac{Y}{x}=\dfrac{7}{9}\), i.e. \(7:9\).
Final Answer:
The ratio Y to x is 7 : 9. \[ \boxed{\text{(A) }7:9} \]
\(T=2\pi\sqrt{\dfrac{m}{k}}\), so \(T\propto\sqrt{m}\).
Step 2: Set up the ratio
\(\dfrac{T'}{T}=\sqrt{\dfrac{x+Y}{x}}=\dfrac43\).
Step 3: Square
\(\dfrac{x+Y}{x}=\dfrac{16}{9}\Rightarrow1+\dfrac{Y}{x}=\dfrac{16}{9}\).
Step 4: Solve
\(\dfrac{Y}{x}=\dfrac{7}{9}\), i.e. \(7:9\).
Final Answer:
The ratio Y to x is 7 : 9. \[ \boxed{\text{(A) }7:9} \]
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