Question

The maximum velocity of a particle, executing simple harmonic motion with an amplitude $7\;mm$, is $4.4\;m/s$. The period of oscillation is:
A. $100s$
B. $0.01s$
C. $10s$
D. $0.1s$

Answer 1

Simple harmonic motion is a form of a motion in which the displacement of the body from its mean position is directly proportional restoring force. To answer this question we need to use the formula for displacement in simple harmonic motion as a function of time. In the formula, we need to substitute the given values and we will get the answer that we need.

Complete step by step solution:
In simple harmonic motion the displacement as a function of time we get,
$\;x = Asin(\omega t + \phi )$
Here $A$is the amplitude of the
$\omega$ is the angular frequency
$t$ is the time taken for the object in motion to complete one oscillation
Now to get the velocity we need to differentiate the displacement equation.
Therefore differentiating the above equation we get,
Velocity $v = \dfrac{{dx}}{{dt}} = A\omega cos(\omega t + \phi )$
Here velocity will be maximum when ${v_{\max }} = A\omega$
We know that the angular velocity $\omega = \dfrac{{2\pi }}{T}$
Given the maximum velocity of the particle, ${v_{\max }} = 4.4\;m/s$
Amplitude $A = 7\;mm = 7 \times {10^{ - 3}}m$
Substituting all the values in ${v_{\max }} = A\omega$
we get,
$4.4\; = 7 \times {10^{ - 3}} \times \dfrac{{2\pi }}{T}$
$\Rightarrow T = 7 \times {10^{ - 3}} \times \dfrac{{2\pi }}{{4.4\;}}$
$\Rightarrow T = 9.99 \times {10^{ - 3}}$
$\Rightarrow T \simeq 10 \times {10^{ - 3}}$
$\therefore T \simeq 0.01s$
Therefore the correct option is B.

Note:
The angular frequency ω is given by $\omega = \dfrac{{2\pi }}{T}$. The angular frequency is measured in terms of radians per second. If we take the inverse of the period then we will get its frequency $f = \dfrac{1}{T}$ . 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, $\left( {1Hz = \dfrac{1}{s}} \right)$.The quantity $\phi$ is said to be the phase constant. It can be determined using the initial conditions of the motion. If we consider at $t = 0$ the object will have its maximum displacement along the positive x-direction, then $\phi = 0$, if the object has its maximum displacement along the negative x-direction, then $\phi = \pi$. If the particle is moving through its equilibrium position with the maximum velocity at $t = 0$ in the negative x-direction then $\phi = \dfrac{\pi }{2}$. Then the quantity $\left( {\omega t + \phi {\text{ }}} \right)$ is said to be the phase.