A mass x g is suspended from a light spring. | vedclass

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(A) The time period of a mass $m$ attached to a spring of force constant $k$ is given by $T = 2\pi \sqrt{m/k}$.For the initial mass $x$,the period is

Vedclass · Accepted Answer

$7:9$

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(A) The time period of a mass $m$ attached to a spring of force constant $k$ is given by $T = 2\pi \sqrt{m/k}$.For the initial mass $x$,the period is $T = 2\pi \sqrt{x/k}$.When the mass is increased by $Y \ g$,the new mass is $(x + Y)$ and the new period is $T' = 4T/3$.Thus,$T' = 2\pi \sqrt{(x + Y)/k} = 4T/3$.Substituting $T$ from the first equation: $2\pi \sqrt{(x + Y)/k} = (4/3) \cdot 2\pi \sqrt{x/k}$.Squaring both sides: $(x + Y)/k = (16/9) \cdot (x/k)$.Canceling $k$ from both sides: $x + Y = (16/9)x$.Rearranging to find $Y$: $Y = (16/9)x - x = (7/9)x$.Therefore,the ratio $Y/x = 7/9$.

$A$ mass $x \ g$ is suspended from a light spring. It is pulled in a downward direction and released so that the mass performs $S.H.M.$ of period $T$. If the mass is increased by $Y \ g$,the period becomes $4T/3$. The ratio of $Y/x$ is:

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