Answer

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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.

**Complete step by step answer**

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.

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