Question

The time period of oscillation of the block as shown in figure is:

(A) $2 \pi \sqrt{\dfrac{m}{2 k}}$
(B) $\pi \sqrt{\dfrac{m}{k}}$
(C) $4 \pi \sqrt{\dfrac{m}{k}}$
(D) $2 \pi \sqrt{\dfrac{m}{k}}$

Answer 1

Hint: We should know that the time required by a body exhibiting periodic motion to complete one period is known as the time period. The S.I. unit is second. The number of times a body exhibits unique motion (each period) in one second is known as frequency. Oscillation refers to any periodic motion moving at a distance about the equilibrium position and repeating itself over and over for a period of time. Example the oscillation up and down of a spring, the oscillation side by side of a spring. The oscillation swinging side by side of a pendulum.

Complete step by step answer
As seen from the diagram the two springs are in series to each other. So, in this situation the force on each string is the same but the displacement on each string is different.
Thus, we can write that:
$x=x_{1}+x_{2} \quad F_{1}=-k_{1} x_{1}$
Hence the time period is given as:
$\Rightarrow T=2 \pi \sqrt{\dfrac{L}{g_{e f f}}}=2 \pi \sqrt{\dfrac{L}{g+a}} F_{2}=-k_{2} x_{2}$
Thus, we can write that:
$\dfrac{1}{k_{s}}=\dfrac{1}{k_{1}}+\dfrac{1}{k_{2}} \dfrac{F}{k_{1}}=x_{1}$
$\Rightarrow k_{s}=\dfrac{k_{1} k_{2}}{k_{1}+k_{2}} \dfrac{F}{k_{2}}=x_{2}$
Hence the formula of the time period is given as:
$T=2\pi \sqrt{\dfrac{m(2k+2k)}{2k\times 2k}}=2\pi \sqrt{\dfrac{m}{k}}$
So, we can say that the time period of oscillation is given as $2 \pi \sqrt{\dfrac{m}{k}}$.

Hence the correct answer is option D.

Note We can conclude that for something to oscillate, energy needs to move back and forth between two forms. For example, in a pendulum, energy moves between potential energy and kinetic energy. This movement of energy between the two forms is what causes the oscillation. There are many types of electronic oscillators, but they all operate according to the same basic principle: an oscillator always employs a sensitive amplifier whose output is fed back to the input in phase. Thus, the signal regenerates and sustains itself. This is known as positive feedback. A normal mode of an oscillating system is a pattern of motion in which all parts of the system move sinusoidally with the same frequency and with a fixed phase relation. These fixed frequencies of the normal modes of a system are known as its natural frequencies or resonant frequencies.