A mass M is suspended from a light spring. An additional mass m added to it displaces the spring further by a distance x then its time period is :
A. $T=2\pi \sqrt{\dfrac{mg}{x(M+m)}}$
B. $T=2\pi \sqrt{\dfrac{x(M+m)}{mg}}$
C. $T=\pi \sqrt{\dfrac{x(M+m)}{mg}}$
D. $T=2\pi \sqrt{\dfrac{Mx}{mg}}$
Answer
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Hint:Use the formula for the time period of a spring block system and write the expression for the time period when the mass m is added to the system. Then by equating the spring force and the gravitational force find the value of the spring constant in this case.
Formula used:
$T=2\pi \sqrt{\dfrac{m}{k}}$
${{F}_{g}}=mg$
${{F}_{s}}=kx$
Complete step by step answer:
The given system is a spring block system with a constant force acting on the mass attached to the spring. In this case, the constant force is the gravitational force acting on the mass in the downwards direction.
The time period of a spring block system is given as $T=2\pi \sqrt{\dfrac{m}{k}}$, where m is the total mass attached to the spring and k is the spring constant of the spring.
When the mass m is added, the total mass attached to the spring becomes $M+m$.
Therefore, the time period of the given system is equal to $T=2\pi \sqrt{\dfrac{M+m}{k}}$ ….
(i) Let us now find the expression for k.
The gravitational force on a mass m is equal to ${{F}_{g}}=mg$.
Therefore, when the mass m is added to the system, an additional force acts in downwards
direction. As a result, the mass goes down stretching the spring.
It is given that the spring displaces by a distance x when the mass m is added. When a spring is displaced by a distance x, it exerts a force in the opposite direction of the displacement, whose magnitude is equal to the ${{F}_{s}}=kx$.
This means that the spring force will balance the gravitational force.
i.e. ${{F}_{g}}={{F}_{s}}$
$\Rightarrow mg=kx$
$\Rightarrow k=\dfrac{mg}{x}$
Now, substitute the value of k in (i).
$\Rightarrow T=2\pi \sqrt{\dfrac{M+m}{\dfrac{mg}{x}}}$
$\Rightarrow T=2\pi \sqrt{\dfrac{x(M+m)}{mg}}$.
Therefore, we found the time period of the given system. This means that the correct option is B.
Note:Some students may make a mistake while solving the question. They may equate the spring force to the total gravitational force.
i.e. $(M+m)g=kx$.
However, this will be incorrect because the gravitational force ($Mg$) acting on the mass M was already balanced by the spring before the mass m and the system was in equilibrium.
This means that the spring already had some displacement.
Therefore, when the mass m is added, the spring displaces further due to the gravitational force acting on the mass m only.
Formula used:
$T=2\pi \sqrt{\dfrac{m}{k}}$
${{F}_{g}}=mg$
${{F}_{s}}=kx$
Complete step by step answer:
The given system is a spring block system with a constant force acting on the mass attached to the spring. In this case, the constant force is the gravitational force acting on the mass in the downwards direction.
The time period of a spring block system is given as $T=2\pi \sqrt{\dfrac{m}{k}}$, where m is the total mass attached to the spring and k is the spring constant of the spring.
When the mass m is added, the total mass attached to the spring becomes $M+m$.
Therefore, the time period of the given system is equal to $T=2\pi \sqrt{\dfrac{M+m}{k}}$ ….
(i) Let us now find the expression for k.
The gravitational force on a mass m is equal to ${{F}_{g}}=mg$.
Therefore, when the mass m is added to the system, an additional force acts in downwards
direction. As a result, the mass goes down stretching the spring.
It is given that the spring displaces by a distance x when the mass m is added. When a spring is displaced by a distance x, it exerts a force in the opposite direction of the displacement, whose magnitude is equal to the ${{F}_{s}}=kx$.
This means that the spring force will balance the gravitational force.
i.e. ${{F}_{g}}={{F}_{s}}$
$\Rightarrow mg=kx$
$\Rightarrow k=\dfrac{mg}{x}$
Now, substitute the value of k in (i).
$\Rightarrow T=2\pi \sqrt{\dfrac{M+m}{\dfrac{mg}{x}}}$
$\Rightarrow T=2\pi \sqrt{\dfrac{x(M+m)}{mg}}$.
Therefore, we found the time period of the given system. This means that the correct option is B.
Note:Some students may make a mistake while solving the question. They may equate the spring force to the total gravitational force.
i.e. $(M+m)g=kx$.
However, this will be incorrect because the gravitational force ($Mg$) acting on the mass M was already balanced by the spring before the mass m and the system was in equilibrium.
This means that the spring already had some displacement.
Therefore, when the mass m is added, the spring displaces further due to the gravitational force acting on the mass m only.
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