MHT CET · Physics · Laws of Motion
A mass ' \(M\) ' is suspended from a light spring. An additional mass \(\mathrm{M}_1\) added extends the spring further by a distance ' \(\mathrm{x}\) '. Now the combined mass will oscillate on the spring with period \(\mathrm{T}=\)
- A \(2 \pi\left[\left(\frac{M_1 g}{x\left(M+M_1\right)}\right)\right]^{\frac{1}{2}}\)
- B \(2 \pi\left[\frac{\left(\mathrm{M}+\mathrm{M}_1\right) \mathrm{x}}{\mathrm{M}_1 \mathrm{~g}}\right]^{\frac{1}{2}}\)
- C \(\left(\frac{\pi}{2}\right)\left[\left(\frac{M_1 g}{x\left(M+M_1\right)}\right)\right]^{\frac{1}{2}}\)
- D \(2 \pi\left[\frac{\left(\mathrm{M}+\mathrm{M}_1\right) \mathrm{x}}{\mathrm{M}_1 \mathrm{~g}}\right]^{\frac{1}{2}}\)
Answer & Solution
Correct Answer
(B) \(2 \pi\left[\frac{\left(\mathrm{M}+\mathrm{M}_1\right) \mathrm{x}}{\mathrm{M}_1 \mathrm{~g}}\right]^{\frac{1}{2}}\)
Step-by-step Solution
Detailed explanation
Time period \(T=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}\)
Here, \(k x=M_1 g\)
\(\therefore \quad \mathrm{K}=\frac{\mathrm{M}_1 \mathrm{~g}}{\mathrm{x}}\)
\(\therefore \quad\) Combined mass: \(\mathrm{M}+\mathrm{M}_1\)
\(\therefore \quad \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{x}\left(\mathrm{M}+\mathrm{M}_1\right)}{\mathrm{M}_1 \mathrm{~g}}}\)
Here, \(k x=M_1 g\)
\(\therefore \quad \mathrm{K}=\frac{\mathrm{M}_1 \mathrm{~g}}{\mathrm{x}}\)
\(\therefore \quad\) Combined mass: \(\mathrm{M}+\mathrm{M}_1\)
\(\therefore \quad \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{x}\left(\mathrm{M}+\mathrm{M}_1\right)}{\mathrm{M}_1 \mathrm{~g}}}\)
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