Question

If an object oscillates $40$ times in one second. The frequency is:

Answer 1

Oscillation is defined by the movement back and forth with a regular rhythm. In physics, oscillation is the repetitive variation of any quantity ( typically in time) or measure about its equilibrium value.
And, the frequency is the number of one full oscillation in unit time. So, to find the frequency we need the time and the number of oscillations.
Formula used:
Frequency $f = (n \times t)Hz$
Where, $n =$ the number of oscillation and, $t =$ time.

Complete step-by-step solution:
An oscillation occurs when an object moves back and forth in a regular rhythm between two positions. And, frequency is the number of a full oscillation for this object in unit time.
In the above question, the number of oscillations is given $40$ times in one second.
We know, Frequency $f = (n \times t)Hz$
Where, $n =$ the number of oscillation and, $t =$ time.
So, here Frequency $f = (40 \times 1)Hz = 40Hz$
Hence, if an object oscillates $40$ times in one second. The frequency is: $40Hz$(answer)
Additional information:
There are two types of oscillations: 1. Damped oscillation and 2. Underdamped oscillation.
Damped oscillation: Damped oscillation occurs when the restoring force is less than the restraining force.
Damped oscillation is of three types :
1. Underdamped oscillation, 2. Critically damped oscillations and, 3. Overdamped oscillation.
Under-Damped oscillation: Underdamped oscillation occurs when the restoring force is equal to the restraining force.

Note: The frequency $f = \dfrac{1}{T} = \dfrac{\omega }{{2\pi }}$ of the motion gives the number of complete oscillations per unit time. It is measured in units of Hertz. $1Hz = \dfrac{1}{{1{\text{second}}}}$
Where, $T$ is the period, $\omega$ is the angular momentum.
The smallest interval of your time during which the system undergoing oscillator returns to the state it was in at a time every which way chosen because of the starting of the oscillation.