A body is suspended from the two light springs separately. The periods of vertical oscillations are T and 3T respectively. The same body is suspended from two springs connected first in parallel and t

Question

A body is suspended from the two light springs separately. The periods of vertical oscillations are T and 3T respectively. The same body is suspended from two springs connected first in parallel and then in series. The periods of oscillation in both cases, respectively, are

TestHub · Accepted Answer

The time period of a spring-mass system is given by $T = 2 π \frac{m}{k}$ , which implies $k = \frac{4 π ^{2} m}{T ^{2}}$ .

For the first spring:
$k_{1} = \frac{4 π ^{2} m}{T ^{2}}$

For the second spring:
$k_{2} = \frac{4 π ^{2} m}{( 3 T ) ^{2}} = \frac{4 π ^{2} m}{9 T ^{2}}$

When the springs are connected in parallel, the equivalent spring constant is $k_{p} = k_{1} + k_{2}$ .
$\frac{1}{T _{p}^{2}} = \frac{k _{p}}{4 π ^{2} m} = \frac{k _{1} + k _{2}}{4 π ^{2} m} = \frac{1}{T _{1}^{2}} + \frac{1}{T _{2}^{2}}$
$\frac{1}{T _{p}^{2}} = \frac{1}{T ^{2}} + \frac{1}{9 T ^{2}} = \frac{10}{9 T ^{2}}$
$T_{p} = \frac{3 T}{10}$

When the springs are connected in series, the equivalent spring constant is given by $\frac{1}{k _{s}} = \frac{1}{k _{1}} + \frac{1}{k _{2}}$ .
$T_{s}^{2} = \frac{4 π ^{2} m}{k _{s}} = 4 π^{2} m (\frac{1}{k _{1}} + \frac{1}{k _{2}}) = T_{1}^{2} + T_{2}^{2}$
$T_{s}^{2} = T^{2} + (3 T)^{2} = 10 T^{2}$
$T_{s} = 10 T$

The periods of oscillation in parallel and series are $\frac{3 T}{10}$ and $10 T$ respectively.

Answer: $\frac{3 T}{10}, 10 T$

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