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(A) The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{M}{K}}$,where $K$ is the spring constant.From this,we see that $T \propt

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$\frac{9}{16}$

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(A) The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{M}{K}}$,where $K$ is the spring constant.From this,we see that $T \propto \sqrt{M}$,or $M \propto T^2$.Let the initial mass be $M$ with time period $T_1 = T$.Let the new mass be $M' = M + m$ with time period $T_2 = \frac{5}{4}T$.Taking the ratio of the two states:$\frac{M'}{M} = \left( \frac{T_2}{T_1} \right)^2$$\frac{M + m}{M} = \left( \frac{\frac{5}{4}T}{T} \right)^2$$1 + \frac{m}{M} = \left( \frac{5}{4} \right)^2$$1 + \frac{m}{M} = \frac{25}{16}$$\frac{m}{M} = \frac{25}{16} - 1 = \frac{9}{16}$.

$A$ mass $M$ is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes simple harmonic oscillations with a time period $T$. If the mass is increased by $m$,then the time period becomes $\frac{5}{4}T$. The ratio of $\frac{m}{M}$ is

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