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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 $\mathrm{T}=$
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The correct answer is:
$2 \pi\left[\frac{\left(\mathrm{M}+\mathrm{M}_1\right) \mathrm{x}}{\mathrm{M}_1 \mathrm{~g}}\right]^{\frac{1}{2}}$
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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