Question

$\large \bf \dag {\underline{Question:-}}$

The time period of vertical oscillation of a light spring with a block A of mass M is T.If another block B of mass m is attached to block A ,the time period of oscillations is doubled.Find the value of $\bf {\frac{m}{M}}$

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Answer 1

$\large \rm {\underbrace{\underline{Question\: 1:}}}$

$\bf\red{\underline{\underline{we \: know\: that:}}} \\ \\ \bf \green{\underline{\underline{time\: period\: :}}}$

$\bf= 2\pi \sqrt{ \frac{m}{k} }$

$\bf \blue {{{where \: m\:is \: mass}}}$

$\bf\ \: {{{and}}}$

$\bf \blue {{{where \: k\:is \:spring \: constant}}}$

$\bf \pink {{{according \: to \: question}}}$

$\bf{{T}^{1}=2 \: T}$

$\large \rm{\underbrace{\underline{now}}}$

$\bf\red{{T}^{1}=2 \: T=2\pi \sqrt{ \frac{M+ m}{k} }}$

$\bf\blue{{T}^{1} = \cancel2 \: T} = \cancel2k\sqrt{ \frac{M+ m}{k}}$

$\bf{ \frac{\orange{T}^{2}}{ {\orange\pi}^{2} } = \frac{M+ m}{k} }$

$\bf{{{divide \: by \: m}}}$

$\bf\frac{k \: \: \red{T}^{2} }{ {\pi}^{2} M} = \pink{1 + \frac{M}{m}}$

$\bf\green{ \frac{M}{m}} =\frac{k \: \: \red{T}^{2} }{ {\pi}^{2} M}- 1$

$\large \rm{\underbrace{\underline{put \: }}}$

$\bf\red{T}^{2} = 4 {\pi}^{2}\pink{ \frac{M}{k}}$

$\large \rm{\underbrace{\underline{so \: :}}}$

$\bf \frac{k \: \red{T}^{2} }{ {\pi}^{2} m} = \cancel\frac{k \: \red{T}^{2} }{ {\pi}^{2} m}$

$\ \bf 4 { \cancel\pi}^{2} \green { \cancel\frac{m}{k} }$

$\large \rm{\underbrace{\underline{so \: :}}}$

$\bf\frac{m}{M} = 4 - 1 = 3$

$\bf \purple{Answer \: is \: 3}$

• refer the attachment

Answer 2

weknowthat:

timeperiod:

$\begin{gathered} \bf= 2\pi \sqrt{ \frac{m}{k} } \\ \end{gathered}=2πkm$

$\bf \blue {{{where \: m\:is \: mass}}}$

$\bf\ \: {{{and}}}$

$\bf \blue {{{where \: k\:is \:spring \: constant}}}$

$\bf \pink {{{according \: to \: question}}}$

$\bf{{T}^{1}=2 \: T}T1=2T$

$\large \rm{\underbrace{\underline{now}}}$

$\begin{gathered} \bf\red{{T}^{1}=2 \: T=2\pi \sqrt{ \frac{M+ m}{k} }} \\ \end{gathered}T1=2T=2πkM+m$

$\begin{gathered} \bf\blue{{T}^{1} = \cancel2 \: T} = \cancel2k\sqrt{ \frac{M+ m}{k}} \\ \end{gathered}T1=2T=2kkM+m$

$\begin{gathered}\bf{ \frac{\orange{T}^{2}}{ {\orange\pi}^{2} } = \frac{M+ m}{k} } \\ \end{gathered}π2T2=kM+m$

$\bf{{{divide \: by \: m}}}$

$\begin{gathered} \bf\frac{k \: \: \red{T}^{2} }{ {\pi}^{2} M} = \pink{1 + \frac{M}{m}} \\ \end{gathered}π2MkT2=1+mM$

$\begin{gathered}\bf\green{ \frac{M}{m}} =\frac{k \: \: \red{T}^{2} }{ {\pi}^{2} M}- 1 \\ \end{gathered}mM=π2MkT2−1$

$\large \rm{\underbrace{\underline{put \: }}}$

$\begin{gathered} \bf\red{T}^{2} = 4 {\pi}^{2}\pink{ \frac{M}{k}} \\ \end{gathered}T2=4π2kM$

$\large \rm{\underbrace{\underline{so \: :}}}$

$\begin{gathered} \bf \frac{k \: \red{T}^{2} }{ {\pi}^{2} m} = \cancel\frac{k \: \red{T}^{2} }{ {\pi}^{2} m} \\ \end{gathered}π2mkT2=π2mkT2<$

so:

4−1=3