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Important Terms In Simple Harmonic Motion MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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Terms associated with SHM is considered one of the most asked concept.
21 Questions around this concept.

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Two simple harmonic motions are represented by the equations $y_{1}= 0.1\sin \left ( 100\pi t+\frac{\pi }{3} \right )\: and \: y_{2}= 0.1\cos \pi t.$

The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 at t=0 is:

$-\pi /3$

$\pi /6$

$-\pi /6$

$\pi /3$

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A point mass oscillates along the x-axis according to law $x= x_{0}\cos \left ( \omega t-\pi /4 \right )$ . If the acceleration of the particle is written as $a= A\cos \left ( \omega t+\delta \right )$ then

$A=x_{0}\omega ^{2},\delta = 3\pi/4$

$A= x_{0},\delta = -\pi /4$

$A=x_{0}\omega ^{2},\delta = \pi/4$

$A= x_{0}\omega ^{2},\delta = -\pi /4$

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Concepts Covered - 1

Terms associated with SHM

Amplitude:-

We know that displacement of a particle in SHM is given by:

$x=A Sin(\omega t+\phi )$

The quantity A is called the amplitude of the motion. It is a positive constant which represents the magnitude of the maximum displacement of the particle from mean position in either direction.

Time period:-

In SHM, a particle repeats its motion after a fixed interval of time. And this time interval after which the particle repeats its motion is called time period. It is denoted by T.
Time period is also defined as the time taken to complete one oscillation. And after one time period, both displacement and velocity of the particle are repeated.
We know that:-

$x=A Sin(\omega t+\phi )$

If a motion is periodic with a period T, then the displacement x (t) must return to its initial value after one period of the motion; that is, x (t) must be equal to x (t + T ) for all t and velocity v(t) must also return to its initial value, i.e., v(t) must be equal to v(t+T). So,

$x(t)=x(t+T)$

$\Rightarrow A Sin(\omega t+\phi)=ASin[\omega [t+T]+\phi]$

$\Rightarrow Sin(\omega t+\phi)=Sin[\omega [t+T]+\phi]$

And

$v(t)=v(t+T)$

$\Rightarrow A \omega Cos(\omega t+\phi)=A\omega Cos[\omega [t+T]+\phi]$

$\Rightarrow Cos(\omega t+\phi)=Cos[\omega [t+T]+\phi]$

As we know both Sine and Cosine function repeats itself when their argument increases by 2π,i.e.,

$\omega t+\phi+2\pi=\omega (t+T)+\phi$

$\Rightarrow 2\pi=T\omega$

$\\\Rightarrow T=\frac{2\pi}{\omega}=2\pi \sqrt\frac{m}{k}; \\ where\ k=force\ or\ spring\ constant\ and\ m=mass$

Time period can also be written as:-

$T=\frac{2\pi}{\omega}=2\pi \sqrt\frac{m}{k}=2\pi \sqrt \frac{m}{\frac{Force}{displacement}}=2\pi \sqrt \frac{m \times displacement}{m \times acceleration}$

$\Rightarrow T=2\pi \sqrt \frac{displacement}{acceleration}$

Frequency:-

The reciprocal of T gives the number of repetitions that occur per unit time. This quantity is called the frequency of the periodic motion.

It is denoted by f.

$f=\frac{1}{T}=\frac{\omega}{2\pi}=\frac{1}{2\pi} \sqrt \frac{k}{m}$

$\Rightarrow \omega=2\pi f;\ where\ \omega\ is\ angular\ frequency$

The unit of frequency is or Hertz(Hz).

Phase:-

$The\ quantity\ (\omega t+\Delta \phi)\ is\ called\ the\ phase.$
It determines the status of the particle in simple harmonic motion.
If the phase is zero at a certain instant, then:

$x=A\ Sin(\omega t+\phi)=0\ and\ v=A\omega \ Cos(\omega t+\phi)=A\omega$

which means that the particle is crossing the mean position and is going towards the positive direction.

Fig:- Status of the particle at different phases

Phase constant:-

The constant $\phi$ is called the phase constant (or phase angle).
The value of $\phi$ depends on the displacement and velocity of the particle at t = 0 or we can say the phase constant signifies the initial conditions.
Any instant can be chosen as t = 0 and hence the phase constant can be chosen arbitrarily.