Question

Let \[{{T}_{1}}\] and \[{{T}_{2}}\] be the time periods of springs \[A\] and \[B\] when mass \[M\] is suspended from one end of each spring. If both springs are taken in series and the same mass \[M\] is suspended from the series combination, the time period is \[T\], then:
A. \[{{T}_{1}}+{{T}_{2}}+{{T}_{3}}\]
B. \[\dfrac{1}{T}=\dfrac{1}{{{T}_{1}}}+\dfrac{1}{{{T}_{2}}}\]
C. \[{{T}^{2}}={{T}_{1}}^{2}+{{T}_{2}}^{2}\]
D. \[\dfrac{1}{{{T}^{2}}}=\dfrac{1}{{{T}_{1}}^{2}}+\dfrac{1}{{{T}_{2}}^{2}}\]

Answer 1

Hint: Recall the formula for time period of oscillation \[\left( T \right)\] of a spring i.e., \[T=2\pi \sqrt{\dfrac{M}{k}}\]. Where,

\[T\] = time period
\[M\] = mass of the block
\[k\] = spring constant

and the formula for series combination of springs,

\[\dfrac{1}{{{k}_{net}}}=\dfrac{1}{{{k}_{1}}}+\dfrac{1}{{{k}_{2}}}\].

Where symbols have their usual meaning.

Complete step by step answer:
Mathematically, time period of oscillation for the spring is given by,

\[T=2\pi \sqrt{\dfrac{M}{k}}\]

Where symbols have their usual meaning.

Now, on squaring both sides, spring constant can be given by
\[\Rightarrow k=\dfrac{4{{\pi }^{2}}M}{{{T}^{2}}}\]
In the same manner the spring constant for first spring will be,
\[{{k}_{1}}=\dfrac{4{{\pi }^{2}}M}{{{T}_{1}}^{2}}\] --------(1)
and the spring constant for second spring will be,
\[{{k}_{2}}=\dfrac{4{{\pi }^{2}}M}{{{T}_{2}}^{2}}\] --------(2)
As we know that for series combination of springs,
\[\dfrac{1}{{{k}_{net}}}=\dfrac{1}{{{k}_{1}}}+\dfrac{1}{{{k}_{2}}}\] --------(3)

Let the time period of oscillation for the series combination of springs is \[T\].

On putting values of \[{{k}_{1}}\] and \[{{k}_{2}}\] from equation (1) and (2), in equation (3) we get

\[\Rightarrow \dfrac{{{T}^{2}}}{4{{\pi }^{2}}M}=\dfrac{{{T}_{1}}^{2}}{4{{\pi }^{2}}M}=\dfrac{{{T}_{2}}^{2}}{4{{\pi }^{2}}M}\]

On solving, we have

\[\Rightarrow {{T}^{2}}={{T}_{1}}^{2}+{{T}_{2}}^{2}\]

So, the relation between time periods, of series combination of spring and for separate springs will be

\[{{T}^{2}}={{T}_{1}}^{2}+{{T}_{2}}^{2}\]
Hence, the correct option is C, i.e., \[{{T}^{2}}={{T}_{1}}^{2}+{{T}_{2}}^{2}\].

Note: Students need to memorize the formula for time period \[\left( T \right)\] of oscillation and they also need to know that how to find resultant spring constant \[\left( {{k}_{net}} \right)\] when springs are connected in series. Students should be very careful while doing calculations.