Question

A particle at the end of a spring executes S.H.M with a period $t_1$, while the corresponding period for another spring is $t_2$. If the period of oscillation with the two springs in series is $T$ then

• A. $T^{-1}=t_1^{-1}+t_2^{-1}$
• B. $T^2=t_1^2+t_2^2$
• C. $T=t_1+t_2$
• D. $T^{-2}=t_1^{-2}+t_2^{-2}$

Answer 1

The correct option is B

$T^2=t_1^2+t_2^2$

For first spring, $t_1=2\pi \sqrt{\frac{m}{k_1}}\Rightarrow k_1=\frac{{\left(2\pi \right)}^{2}m}{{\left(t_1\right)}^2}$

$t_2=2\pi \sqrt{\frac{m}{k_2}}\Rightarrow k_2=\frac{{\left(2\pi \right)}^{2}m}{{\left(t_2\right)}^2}$

When the spring are in series then,
$\frac{1}{k_{eff}}=\frac{1}{k_1}+\frac{1}{k_2}$

$\frac{1}{k_{eff}}=\frac{k_1+k_2}{k_1{k}_2}$

$k_{eff}=\frac{k_1{k}_2}{k_1+k_2}$

Time period when spring is in series combination

$T=2\pi \sqrt{\frac{m(k_1+k_2)}{k_1{k}_2}}$

$T=2\pi \sqrt{\frac{m}{k_2}+\frac{m}{k_1}}$

$=2\pi \sqrt{\frac{{\left(t_2\right)}^2}{{\left(2\pi \right)}^2}+\frac{{\left(t_1\right)}^2}{{\left(2\pi \right)}^2}}$

$\frac{T^2}{{\left(2\pi \right)}^2}=\frac{{\left(t_2\right)}^2}{{\left(2\pi \right)}^2}+\frac{{\left(t_1\right)}^2}{{\left(2\pi \right)}^2}$

$T^2=t_1^2+t_2^2$