Question

3. compare the time period of
two simple pendulum of length
1m and 16m at a place​

Answer 1

Answer:

$\green{\tt{\therefore{T_{1}:T_{2}=1:4}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\begin{gathered}\green{\underline \bold{Given :}} \\ \tt:\implies Length( l_{1}) = 1 \: m \\ \\ \tt:\implies Length( l_{2}) = 16\: m \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Comparison \:of \: time \: period =?\end{gathered}$

• According to given question :

$\begin{gathered}\bold{As \: we \: know \: that} \\ \tt: \implies T_{1} = 2\pi \sqrt{ \frac{ l_{1} }{g} } \\ \\ \tt: \implies T_{1} =2\pi \sqrt{ \frac{1}{g} } - - - - - (1) \\ \\ \bold{Similarly : } \\ \tt: \implies T_{2} = 2\pi \sqrt{ \frac{ l_{2}}{g} } \\ \\ \tt: \implies T_{2} = 2\pi \sqrt{ \frac{16}{g} } - - - - - (2) \\ \\ \text{Taking \: ratio \: of \: (1) \: and \: (2)} \\ \\ \tt: \implies \frac{ T_{1} }{ T_{2} } = \frac{ 2\pi\sqrt{ \frac{1}{g} } }{ 2\pi \sqrt{ \frac{16}{g} } } \\ \\ \tt: \implies \frac{ T_{1} }{ T_{2} } = \sqrt{ \frac{1}{g} } \times \sqrt{ \frac{g}{16} } \\ \\ \tt: \implies \frac{ T_{1} }{ T_{2} } = \sqrt{ \frac{1}{16} } \\ \\ \tt: \implies \frac{ T_{1} }{ T_{2} } = \frac{1}{4} \\ \\ \green{\tt: \implies { T_{1} } : { T_{2} } =1 : 4} \\ \\ \green{\tt \therefore Time \: period \: of \: l_{2} \: is \:4 \: times \: of \: time \: \: period \: \: of \: \: l_{1}}\end {gathered}$