Question

A system is shown in the figure. The time period for small oscillations of the two identical blocks will be

A. $2\pi \sqrt {\dfrac{m}{k}} $
B. $2\pi \sqrt {\dfrac{m}{{2k}}} $
C. $2\pi \sqrt {\dfrac{m}{{4k}}} $
D. $2\pi \sqrt {\dfrac{m}{{8k}}} $

Answer 1

Hint:The time period is defined as the time taken for the completion of one oscillation. In order to calculate the time period of the oscillation of the system, we have to calculate the angular frequency of the system, denoted by $\omega $, which can be done by calculating the net force acting on the system.

Complete step-by-step answer:
The spring-mass system exhibits simple harmonic motion. For a simple harmonic motion, the restoring force at any point is directly proportional to the displacement of the body at that particular point.
Restoring force, $F \propto - x$
The negative sign indicates the direction of the force in relation to the displacement.
Consider the following spring-mass system as shown:

Now, let us move the mass slightly towards the right direction. Then, the spring on the right gets compressed while the spring on the left gets stretched. Now, there are two restoring forces ${F_1}$ and ${F_2}$ acting on the mass m, as shown below.

Here, the ${F_1}$ is the restoring force acting due to the compression of the spring on the right and ${F_2}$ is the restoring force acting due to the stretching of the spring on the left side.
The restoring forces,
${F_1} = - kx$
${F_2} = - kx$
From the above figure, the net force, $F = {F_1} + {F_2}$
$F = {F_1} + {F_2} = - kx - kx = - 2kx$
The force applied, $F = ma$ where m is the mass and a is the acceleration of the system.
Since, this is a simple harmonic motion, we have the relation between acceleration and displacement as–
$a = - {\omega ^2}x$
where $\omega $ is called the angular frequency of the system.
To calculate $\omega $ –
Substituting the acceleration in the equation for force –
$F = ma$
$ - 2k = - m{\omega ^2}x$
$ \Rightarrow {\omega ^2} = \dfrac{{2k}}{m}$
$ \Rightarrow \omega = \sqrt {\dfrac{{2k}}{m}} $
The angular frequency, $\omega = 2\pi f$ where f is the frequency.
Since, frequency is the inverse of the time period T, we have –
$\omega = \dfrac{{2\pi }}{T}$
Substituting for $\omega $ –
$\dfrac{{2\pi }}{T} = \sqrt {\dfrac{{2k}}{m}} $
Solving for T –
$T = 2\pi \sqrt {\dfrac{m}{{2k}}} $

Hence, the correct option is Option B.

Note:There is an alternative approach to find the net force acting on the mass.
Here, the springs are in a parallel combination. When the springs are in a parallel combination, the effective spring constant is equal to the sum of the spring constants.
$K = {K_1} + {K_2}$
$K = k + k = 2k$
The restoring force, $F = - Kx$
Thus, $F = - 2kx$
We can get the value of the net force directly, by this method.