Q118 / 150 · MHT CET 2023 (13 May Shift 2) · Physics

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
    $2 \pi\left[\left(\frac{M_1 g}{x\left(M+M_1\right)}\right)\right]^{\frac{1}{2}}$
  • Correct answer
    B
    $2 \pi\left[\frac{\left(\mathrm{M}+\mathrm{M}_1\right) \mathrm{x}}{\mathrm{M}_1 \mathrm{~g}}\right]^{\frac{1}{2}}$
  • C
    $\left(\frac{\pi}{2}\right)\left[\left(\frac{M_1 g}{x\left(M+M_1\right)}\right)\right]^{\frac{1}{2}}$
  • D
    $2 \pi\left[\frac{\left(\mathrm{M}+\mathrm{M}_1\right) \mathrm{x}}{\mathrm{M}_1 \mathrm{~g}}\right]^{\frac{1}{2}}$

Solution

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: $\mathrm{M}+\mathrm{M}_1$ $\therefore \quad \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{x}\left(\mathrm{M}+\mathrm{M}_1\right)}{\mathrm{M}_1 \mathrm{~g}}}$

Hence, the correct answer is option: 'B'.

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