Question

Two dissimilar spring fixed at one end are stretched by 10cm and 20cm respectively, when mass ${{m}_{1}}$ and ${{m}_{2}}$ are suspended at their lower ends. When displaced slightly from their mean positions and released, they will oscillate with period in the ratio
A. $1 : 2$
B. $2 : 1$
C. $1 : 1.41$
D. $1.41 : 4$

Answer 1

In this case, the gravitational force acting on the masses will be equal to the spring force acting on each of the masses. Equate the two forces for both the spring mass system and use the formula for the time period of a hanging spring mass system.

Formula used:
$T=2\pi \sqrt{\dfrac{m}{k}}$
${{F}_{g}}=mg$
${{F}_{s}}=kx$
Where $k$ is the spring constant of the spring, $g$ is acceleration due to gravity, $m$ is the mass of the spring and $x$ be the extension in the spring.

Complete step by step answer:
The time period of a hanging spring block system is given to be equal to $T=2\pi \sqrt{\dfrac{m}{k}}$, where m is the mass of the block and k is the spring constant of the spring. It is said that when the masses ${{m}_{1}}$ and ${{m}_{2}}$ are suspended at the lower ends of the two springs, the springs are displaced by 10 cm and 20 cm respectively.
This is due to the gravitational force acting (in the downward direction) on the two masses. When the springs are displaced by the said amounts, the spring force exerted by the spring is equal in magnitude but opposite in direction to the gravitational force.
i.e. ${{F}_{g}}={{F}_{s}}$.
The gravitational force acting on a mass m is given as ${{F}_{g}}=mg$, where g is acceleration due to gravity. When the spring is stretched by a length x, then the spring force acting on the mass is equal to ${{F}_{s}}=kx$.
Therefore, the gravitational force on mass ${{m}_{1}}$ is equal to ${{F}_{g,1}}={{m}_{1}}g$.
And the spring force acting on this mass is ${{F}_{s,1}}={{k}_{1}}{{x}_{1}}={{k}_{1}}(10)$.
But we know that ${{F}_{g,1}}={{F}_{s,1}}$.
$\Rightarrow {{m}_{1}}g=10{{k}_{1}}$
$\Rightarrow {{k}_{1}}=\dfrac{{{m}_{1}}g}{10}$.
Therefore, the time period of the first spring is,
${{T}_{1}}=2\pi \sqrt{\dfrac{{{m}_{1}}}{{{k}_{1}}}}\\\ \Rightarrow {{T}_{1}} =2\pi \sqrt{\dfrac{{{m}_{1}}}{\dfrac{{{m}_{1}}g}{10}}}$.
$\Rightarrow {{T}_{1}}=2\pi \sqrt{\dfrac{10}{g}}$ ….. (i)
Similarly, the gravitational force on the mass ${{m}_{1}}$ is equal to ${{F}_{g,2}}={{m}_{2}}g$.
And the spring force acting on this mass is ${{F}_{s,2}}={{k}_{2}}{{x}_{2}}={{k}_{2}}(20)$.
But we know that ${{F}_{g,2}}={{F}_{s,2}}$.
$\Rightarrow {{m}_{2}}g=20{{k}_{2}}$
$\Rightarrow {{k}_{2}}=\dfrac{{{m}_{2}}g}{20}$.
Therefore, the time period of the second spring is ${{T}_{2}}=2\pi \sqrt{\dfrac{{{m}_{2}}}{{{k}_{2}}}}=2\pi \sqrt{\dfrac{{{m}_{2}}}{\dfrac{{{m}_{2}}g}{20}}}$.
$\Rightarrow {{T}_{2}}=2\pi \sqrt{\dfrac{20}{g}}$ …. (ii).
Now, divide (i) by (ii).
$\dfrac{{{T}_{1}}}{{{T}_{2}}}=\dfrac{2\pi \sqrt{\dfrac{10}{g}}}{2\pi \sqrt{\dfrac{20}{g}}}$
$\Rightarrow \dfrac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\dfrac{10}{20}}\\\ \Rightarrow \dfrac{{{T}_{1}}}{{{T}_{2}}} =\dfrac{1}{\sqrt{2}}\\\ \therefore \dfrac{{{T}_{1}}}{{{T}_{2}}} =\dfrac{1}{1.41}$.
This means that the ratio of the time period of the first spring to the second spring is equal to 1 : 1.41.

Hence, the correct option is C.

Note: There can be some misconception that the time period of oscillations of a spring block system will depend on the amplitude of the oscillations. However, the time period is independent of the amplitude of the oscillations. This means that for a given spring block system, the time period remains the same for every value of amplitude.