The main purpose of this question is to express displacement as a function of time when an object is moving in a Simple Harmonic Motion.

The Simple Harmonic Motion is a repeated back-and-forth movement through a central position or equilibrium such that on one side of this position the maximum displacement equals the maximum displacement on another side. Every entire vibration has the same period. Simple Harmonic Motion, which is characterized by the mass oscillation on a spring when subjected to the linear elastic force applied offered by Hooke’s law, can represent a mathematical model for a wide range of movements. The movement is periodic in time and has only one resonant frequency.

All the Simple Harmonic Motions are repetitive and periodic, but all the oscillatory motions are not simple harmonic. Oscillatory movement is also referred to as the harmonic motion of all the oscillatory movements, the most significant of which is Simple Harmonic Motion. Simple Harmonic Motion is a highly helpful tool for comprehending the attributes of light waves, alternating currents, and sound waves.

## Expert Answer

The object is moving in a positive direction with displacement $-7\,cm$ at time $t=0\,s$. Now, consider the negative cosine function as the object is at the lowest point initially. Generally, displacement as a function of time can be expressed as:

$d=-A\cos(Bt-C)+D$

Let $A$ be the amplitude then $A=7\,cm$ and $T$ be the period of the object then $T=5\,s$. And so:

$T=\dfrac{2\pi}{B}$

$5=\dfrac{2\pi}{B}$

$B=\dfrac{2\pi}{5}$

Let $C$ be the phase shift then $C=0$, since no phase shift exists at $t=0$. Also, let $D$ be the vertical phase shift then $D=0$.

Finally, we can express the displacement $(d)$ as a function of time $(t)$ as follows:

$d=-7\cos\left(\dfrac{2\pi}{5} t-0\right)+0$

$d=-7\cos\left(\dfrac{2\pi t}{5}\right)$

## Example

The time of an object carrying out Simple Harmonic Motion is $3\,s$. Find out the time interval from $t=0$ after which its displacement will be $\dfrac{1}{2}$ of its amplitude.

### Solution

Let $T$ be the period, then:

$T=2\,s$

Let $d$ be the displacement and $A$ be the amplitude, then:

$d=\dfrac{1}{2}A$

Since the particle passes through the mean position, therefore $\alpha=0$.

Let $\omega $ be the angular velocity, then:

$\omega=\dfrac{2\pi}{T}=\dfrac{2\pi}{3}\,rad/s$

Also, the displacement of the object carrying Simple Harmonic Motion is given by:

$d=A\sin(\omega t+\alpha)$

$\dfrac{1}{2}A=A\sin\left(\dfrac{2\pi}{3}t+0\right)$

$\dfrac{1}{2}=\sin\left(\dfrac{2\pi}{3}t\right)$

$\dfrac{2\pi}{3}t=\sin^{-1}\left(\dfrac{1}{2}\right)$

$\dfrac{2\pi}{3}t=\dfrac{\pi}{6}$

$t=\dfrac{1}{4}\,s$