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# The period of oscillation of a mass suspended from a spring is 2 second along with it another mass of $2 \mathrm{kg}$ is suspended, the period of oscillation is by one second. The mass m will be:(A) $2 \mathrm{kg}$ (B) $1 \mathrm{kg}$ (C) $1.6 \mathrm{kg}$ (D) $2.6\text{kg}$

Last updated date: 13th Jun 2024
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Hint
We know that 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. We can write the example as 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. By one oscillation we mean that it is of a simple pendulum is one complete cycle of swinging one way and then returning to its original starting position. Based on this concept we have to solve this question.

Let us assume that,
Mass = m
Time period = T
Spring constant = k
We get,
$T=2\pi \sqrt{\dfrac{m}{k}}$
$\pi$ is a constant $k$ is a constant for a given spring
$\therefore T \propto \sqrt{m}$
From this we get,
$\therefore T \propto \sqrt{m}$
$2s+1s\propto \sqrt{m+2}$
$\therefore \quad \dfrac{3}{2}=\dfrac{\sqrt{m+2}}{\sqrt{m}}$
Now,
$\dfrac{3}{2}=\sqrt{\dfrac{m+2}{m}}$
$\Rightarrow {{\left( \dfrac{3}{2} \right)}^{2}}=\dfrac{9}{4}=\dfrac{m+2}{m}$
From the above equation we get,
$9 m=4(m+2)$
$9 m=4 m+8$
$5m=8$
$m=1.6\text{kg}$

Therefore, the correct answer is Option (C).

Note
We know that the frequency of oscillation is the number of full oscillations in a one-time unit, say in a second. A pendulum that takes 0.5 seconds to make one full oscillation has a frequency of 1 oscillation per 0,5 second, or we can say 2 oscillations per second.
We know that 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.