Question

A uniform spring has certain mass suspended from it and it's period of vertical oscillations is $t_1$. The spring is now cut in $2$ parts having lengths in ratio $1:2$ and these springs are now connected in series and then in parallel. find out the ratio of the time period of these two ossillation?

• A. $1$
• B. $\sin \theta$
• C. $\sqrt{\frac{2}{9}}$
• D. $\sqrt{\frac{9}{2}}$

Answer 1

Answer: (C)

$\sqrt{\frac{2}{9}}$

Let $k$ be initial force constant of spring,$k_1$ and $k_2$ be the force constant of new springs

Since

$kx=constant$

$\Rightarrow \frac{x_1}{x_2}=1/2$

$\Rightarrow \frac{k_1}{k_2}=2$ ........$\left(1\right)$

therefore

$k_1=3k$

$k_2=3k/2$

As initially these lengths were in series:

$\frac{1}{k}=\frac{1}{k_1}+\frac{1}{k_2}$

$\Rightarrow \frac{1}{k_1^{\prime }}=\frac{1}{3k_1}+\frac{2}{3k_1}$

$\Rightarrow \frac{1}{k_1^{\prime }}=\frac{1}{k_1}$

$\Rightarrow {k^{\prime }}_1=k$

When these two stringd are connected in parallel,

${k_2}^{\prime }=k_1+k_2$

${k_2}^{\prime }=\frac{3k}{2}+3k$

${k_2}^{\prime }=\frac{9k}{2}$

Time period is

$\frac{{T_1}^{\prime }}{T_2^{\prime }}=\sqrt{\frac{k_1^{\prime }}{{k_2^{\prime }}^{\prime }}}$

$\Rightarrow \frac{{T_1}^{\prime }}{T_2^{\prime }}=\sqrt{\frac{2}{9}}$