All the springs in fig. (a), (b) and (c) are identical, each having force constant K. Mass attached to each system is ' $m$ '. If $T_{a}, T_{b}$ and $T_{c}$ are the time periods of oscillations of the three systems respectively, then

Question

TestHub · Accepted Answer

For figure (a), the system consists of a single spring of force constant $K$ .

The time period is given by:

$T_{a} = 2 π \frac{m}{K}$

For figure (b), two identical springs are connected in series. The equivalent force constant $K_{b}$ is:

$\frac{1}{K _{b}} = \frac{1}{K} + \frac{1}{K} \Rightarrow K_{b} = \frac{K}{2}$

The time period is:

$T_{b} = 2 π \frac{m}{K _{b}} = 2 π \frac{2 m}{K} = 2 T_{a}$

For figure (c), two identical springs are connected in parallel. The equivalent force constant $K_{c}$ is:

$K_{c} = K + K = 2 K$

The time period is:

$T_{c} = 2 π \frac{m}{K _{c}} = 2 π \frac{m}{2 K} = \frac{T _{a}}{2}$

From the expressions for $T_{b}$ and $T_{c}$ , we get:

$\frac{T _{b}}{T _{c}} = \frac{2 T _{a}}{T _{a} / 2} = 2 \times 2 = 2$

$\Rightarrow T_{b} = 2 T_{c}$

Answer: $T_{b} = 2 T_{c}$

Physics - SHM Question with Solution | TestHub

Options:

Doubts & Discussion