A simple harmonic oscillator having four identical springs (as shown in figure) has a time period $T_{1}$ . Another similar simple harmonic oscillator of the same mass $m$ but the four identical springs with spring constants $2 k$ has a time period $T_{2}$ , then $\frac{T _{1}}{T _{2}}$ is

Question

TestHub · Accepted Answer

The time period of a simple harmonic oscillator is given by $T = 2 π \frac{m}{k _{e q}}$ , where $k_{e q}$ is the equivalent spring constant of the system.

For a given geometric arrangement of springs, the equivalent spring constant is directly proportional to the individual spring constants. Let the equivalent spring constant of the first arrangement (with springs of constant $k$ ) be $k_{e q 1} = C k$ , where $C$ is a constant depending on the geometry of the arrangement.

The time period of the first oscillator is:
$T_{1} = 2 π \frac{m}{C k}$

For the second oscillator, the arrangement is similar but each spring has a spring constant of $2 k$ . The new equivalent spring constant will be:
$k_{e q 2} = C (2 k) = 2 C k$

The time period of the second oscillator is:
$T_{2} = 2 π \frac{m}{2 C k}$

Taking the ratio of the two time periods:
$\frac{T _{1}}{T _{2}} = \frac{2 π \frac{m}{C k}}{2 π \frac{m}{2 C k}} = 2$

Thus, the ratio $T_{1} : T_{2}$ is $2 : 1$ .

Answer: $2 : 1$

Physics - SHM Question with Solution | TestHub

Options:

Doubts & Discussion