Question

A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find acceleration and velocity of the body when the displacement is (a) 5 cm (b) 3 cm (c) 0 cm

Answer 1

The simple harmonic motion is defined as a special type of periodic motion where the restoring force on the moving object is directly proportional, at every instant, to the displacement of the body.
Restoring force, F –
$F \propto - x \\\ F = - kx \\\$
k = constant.

Complete step by step answer:
Newton’s second law states the definition of force as –
$F = ma$
Acceleration is obtained by differentiating the displacement function twice.
We know that, $a = \dfrac{{{d^2}x}}{{d{t^2}}}$
Substituting in the equation of harmonic function –
$F = - kx \\\ \Rightarrow m\left( {\dfrac{{{d^2}x}}{{d{t^2}}}} \right) = - kx \\\ \Rightarrow \dfrac{{{d^2}x}}{{d{t^2}}} = - \dfrac{k}{m}x \\\$
On solving the differential equation, we have the solution –
$x(t) = {c_1}\cos \left( {\omega t} \right) + {c_2}\sin \left( {\omega t} \right)$ where $\omega = \sqrt {\dfrac{k}{m}}$
Applying the initial condition: At $t = 0,x(t) = 0$
$\Rightarrow x(t) = {c_1}\cos \left( 0 \right) + {c_2}\sin \left( 0 \right) \\\ \Rightarrow 0 = {c_1} + 0 \\\ \Rightarrow {c_1} = 0$
Therefore,
$x(t) = A\sin \omega t$ with ${c_2} = A$, called the amplitude which means the maximum displacement about the mean position
On differentiating the equation once, we get the equation for velocity,
$V = \dfrac{{dx}}{{dt}} = \dfrac{d}{{dx}}\left( {A\sin \omega t} \right) \\\ \Rightarrow V = A\omega \cos \omega t$
In trigonometry,
$\Rightarrow {\sin ^2}\theta + {\cos ^2}\theta = 1 \\\ \Rightarrow \cos \theta = \sqrt {1 - {{\sin }^2}\theta }$
Applying this identity in the above equation, we have –
$V = A\omega \cos \omega t \\\ \Rightarrow V = A\omega \sqrt {1 - {{\sin }^2}\omega t}$
Taking the equation inside the square root,
$V = \sqrt {{A^2}{\omega ^2} - {A^2}{\omega ^2}{{\sin }^2}\omega t} \\\ \Rightarrow V = \sqrt {{A^2}{\omega ^2} - {\omega ^2}{x^2}} \because x = A\sin \omega t\& {x^2} = {A^2}{\sin ^2}\omega t \\\ \Rightarrow V = \sqrt {{\omega ^2}\left( {{A^2} - {x^2}} \right)} \\\ \Rightarrow V = \omega \sqrt {\left( {{A^2} - {x^2}} \right)}$
On differentiating the equation again, we get the equation for acceleration,
$a = \dfrac{{{d^2}x}}{{d{t^2}}} = \dfrac{d}{{dt}}\left( {\dfrac{{dx}}{{dt}}} \right) = \dfrac{d}{{dt}}\left( V \right)$
$V = A\omega \cos \omega t$
Substituting and differentiating, we get –

$$a = \dfrac{{dV}}{{dt}} = \dfrac{d}{{dt}}\left( {A\omega \cos \omega t} \right) \\\ \Rightarrow a = - A{\omega ^2}\sin \omega t \\\ \Rightarrow a = - {\omega ^2}x\because x = A\sin \omega t$$

In this problem,
Amplitude, $A = 5cm = 0.05m$
Time period, $T = 0.2s$
The angular velocity, $\omega$ is equal to the number of radians covered per unit time. Thus,
$\Rightarrow \omega = \dfrac{{2\pi }}{T} \\\ Here, \\\ \Rightarrow \omega = \dfrac{{2\pi }}{{0.2}} = 10\pi$
Now, let us consider each case for the displacements –
Case 1: When displacement is 5 cm
$x(t) = 5cm$
Velocity, $V = \dfrac{{dx}}{{dt}} = \omega \sqrt {{A^2} - {x^2}}$
Substituting,
$\Rightarrow V = \omega \sqrt {{A^2} - {x^2}} \\\ \Rightarrow V = 10\pi \sqrt {{{0.05}^2} - {{0.05}^2}} \\\ \Rightarrow V = 10\pi (0) = 0m{s^{ - 1}} \\\$
Acceleration, $A = \dfrac{{{d^2}x}}{{d{t^2}}} = - {\omega ^2}x$
Substituting,
$\Rightarrow A = - {\omega ^2}x \\\ \Rightarrow A = - {\left( {10\pi } \right)^2} \times 0.05 \\\ \Rightarrow A = - 100 \times {\left( {3.14} \right)^2} \times 0.05 = - 49.298m{s^{ - 2}} \\\$
Thus,
$Velocity,V = 0m{s^{ - 1}} \\\ Acceleration,a = - 49.298m{s^{ - 2}} \\\$

Case 2: When displacement is 3 cm
$x(t) = 3cm$
Velocity, $V = \dfrac{{dx}}{{dt}} = \omega \sqrt {{A^2} - {x^2}}$
Substituting,
$V = \omega \sqrt {{A^2} - {x^2}} \\\ \Rightarrow V = 10\pi \sqrt {{{0.05}^2} - {{0.03}^2}} \\\ \Rightarrow V = 10\pi \sqrt {0.0025 - 0.0009} \\\ \Rightarrow V = 10\pi \sqrt {0.0016} \\\ \Rightarrow V = 10\pi \times 0.04 = 1.256m{s^{ - 1}} \\\$
Acceleration, $A = \dfrac{{{d^2}x}}{{d{t^2}}} = - {\omega ^2}x$
Substituting,
$A = - {\omega ^2}x \\\ \Rightarrow A = - {\left( {10\pi } \right)^2} \times 0.03 \\\ \Rightarrow A = - 100 \times {\left( {3.14} \right)^2} \times 0.03 = - 29.578m{s^{ - 2}} \\\$
Thus,
$Velocity,V = 1.256m{s^{ - 1}} \\\ Acceleration,a = - 29.578m{s^{ - 2}} \\\$
Case 3: When displacement is 0 cm
$x(t) = 0cm$
Velocity, $V = \dfrac{{dx}}{{dt}} = \omega \sqrt {{A^2} - {x^2}}$
Substituting,
$V = \omega \sqrt {{A^2} - {x^2}} \\\ \Rightarrow V = 10\pi \sqrt {{{0.05}^2} - 0} \\\ \Rightarrow V = 10\pi \sqrt {0.0025} \\\ \Rightarrow V = 10\pi \times 0.05 = 1.57m{s^{ - 1}} \\\$
Acceleration, $A = \dfrac{{{d^2}x}}{{d{t^2}}} = - {\omega ^2}x$
Substituting,
$A = - {\omega ^2}x \\\ \Rightarrow A = - {\left( {10\pi } \right)^2} \times 0 \\\ \Rightarrow A = 0m{s^{ - 2}} \\\$
Thus,
$Velocity,V = 1.57m{s^{ - 1}} \\\ Acceleration,a = 0m{s^{ - 2}} \\\$

Note: In actual, the equation for simple harmonic motion is –
$x(t) = A\sin \left( {\omega t + \phi } \right)$ where $\phi$ is called the initial phase angle, which represents the angle at which we are starting the vibration.
In general cases, there is a common assumption that the initial phase angle is 0.
Hence, in the question, it is directly taken as:
$x(t) = A\sin \left( {\omega t + 0} \right) = A\sin \omega t$