Question

Consider two identical springs each of spring constant $k$ and negligible mass compared to the mass $M$ as shown. Fig.$1$ shows one of them and Fig. $2$ shows their series combination. The ratio of the time period of oscillation of the two S.H.M is $\frac{T_b}{T_a}=\sqrt{x}$, where the value of $x$ is ______. (Round off to the nearest integer)

Answer 1

Step 1. Given Data:

Spring constant $=k$

The ratio of the time period of oscillation of the two S.H.M is, $\frac{T_b}{T_a}=\sqrt{x}$

Step 2. Finding the spring constant in a series combination

For series combination, the equivalent spring constant, $k_{eq}$

$k_{eq}=\frac{k_1\times k_2}{k_1+k_2}$

Where $k_1$ is the spring constant for the spring $1$and $k_2$ is the spring constant for the spring $2$

As the springs are identical, $k_1=k_2=k$

$k_{eq}=\frac{k}{2}$

Step 3. Finding the ratio of the time period of oscillation of both the S.H.M

By using the formula of the time period,

$T=2\mathrm{\pi }\sqrt{\frac{\mathrm{M}}{\mathrm{k}}}$ [Where $M$ is the mass]

$T_a=2\mathrm{\pi }\sqrt{\frac{\mathrm{M}}{\mathrm{k}}}$

$T_b=2\mathrm{\pi }\sqrt{\frac{\mathrm{M}}{{\displaystyle k_{eq}}}}=2\mathrm{\pi }\sqrt{\frac{2\mathrm{M}}{{\displaystyle k}}}$

The ratio of the time periods

$\frac{T_b}{T_a}=\sqrt{2}$

By comparing the above value to the $\frac{T_b}{T_a}=\sqrt{x}$, we get

Therefore, $\mathit{x}\mathbf{=}\mathbf{2}$