Question

A particle at the end of a spring executes simple harmonic motion with a period \[{t_1}\] while the corresponding period for another spring is ${t_2}$. If the period of oscillation with the two springs in series is $T$, then:
A) $T = {t_1} + {t_2}$
B) ${T^2} = {t_1}^2 + {t_2}^2$
C) ${T^{ - 1}} = {t_1}^{ - 1} + {t_2}^{ - 1}$
D) ${T^{ - 2}} = {t_1}^{ - 2} + {t_2}^{ - 2}$

Answer 1

Hint:It’s easy to solve if we put it this way, the springs are connected in series with a certain time period. We have to compute the value of the spring constant for the combination of springs in terms of time-period and compare this obtained value with the values of spring constant for each individual spring to establish a relation.

Formula used:
For a spring in S.H.M
$T = 2\pi \sqrt {\dfrac{m}{k}} $
Where $T$ is the time period for oscillation
$m$ is the mass of the particle oscillating on the spring
$k$ is the spring constant for the spring in SHM

Complete step by step solution:
Let’s consider two springs A and B who are exhibiting simple harmonic motion.
Now let’s take the mass of the oscillating particle as ${m_{}}$, and the spring constants as ${k_1}$ and ${k_2}$ respectively for spring A and B
For spring A, the time period of oscillation is given by,
${t_1} = 2\pi \sqrt {\dfrac{{{m_{}}}}{{{k_1}}}} $
$ \Rightarrow {k_1} = \dfrac{{4{\pi ^2}{m_{}}}}{{{t_1}^2}}$………. (1)
For the second spring B time period of oscillation is given by,
${t_2} = 2\pi \sqrt {\dfrac{{{m_{}}}}{{{k_2}}}} $
$ \Rightarrow {k_2} = \dfrac{{4{\pi ^2}{m_{}}}}{{{t_2}^2}}$ ……… (2)
For a number of springs connected in series, the collective spring-constant value for all springs is given by the following formula.
$\dfrac{1}{k} = \dfrac{1}{{{k_1}}} + \dfrac{1}{{{k_2}}} + \dfrac{1}{{{k_3}}} + ....$
Where $k$ is the collective spring constant for the springs with individual spring constant values ${k_1},{k_2},{k_3}...$
We are given in the question that the two springs A and B are connected in series, so their collective spring constant ${k_c}$ will be,
$\dfrac{1}{{{k_c}}} = \dfrac{1}{{{k_1}}} + \dfrac{1}{{{k_2}}}$
$ \Rightarrow {k_c} = \dfrac{{{k_1}{k_2}}}{{{k_1} + {k_2}}}$ …….. (3)
So for the two springs connected in series their collective time period would be,
$T = 2\pi \sqrt {\dfrac{m}{{{k_c}}}} $
$ \Rightarrow {k_c} = \dfrac{{4{\pi ^2}m}}{{{T^2}}}$ ………. (4)
Substituting the value of ${k_c}$ we get,
$ \Rightarrow \dfrac{{{k_1}{k_2}}}{{{k_1} + {k_2}}} = \dfrac{{4{\pi ^2}m}}{{{T^2}}}$
Putting the values of ${k_1}$ and ${k_2}$ we get,
$ \Rightarrow \dfrac{{\dfrac{{4{\pi ^2}m}}{{{t_1}^2}} \times \dfrac{{4{\pi ^2}m}}{{{t_2}^2}}}}{{\dfrac{{4{\pi ^2}m}}{{{t_1}^2}} + \dfrac{{4{\pi ^2}m}}{{{t_2}^2}}}} = \dfrac{{4{\pi ^2}m}}{{{T^2}}}$
Simplifying this we get,
$ \Rightarrow \dfrac{{\dfrac{1}{{{t_1}^2{t_2}^2}}}}{{\dfrac{1}{{{t_1}^2}} + \dfrac{1}{{{t_2}^2}}}} = \dfrac{1}{{{T^2}}}$
$ \Rightarrow \dfrac{1}{{{T^2}}} = \dfrac{1}{{{t_1}^2}} + \dfrac{1}{{{t_2}^2}}$
So the relation is ${T^{ - 2}} = {t_1}^{ - 2} + {t_2}^{ - 2}$.

The correct answer is option (D).

Note: When answering these kinds of spring problems check the spring formula for combination of springs twice as a student might get confused between the reciprocal value and the actual value. For resistors in series combination, equivalent resistance is the sum of all resistances but for springs, the reciprocal of the equivalent spring constant is the sum of the reciprocals of all the spring constants for each spring. This is a trick to not get confused between resistors and springs.