Question

A particle executes simple harmonic motion of period $16s$. Two seconds later after it passes through the center of oscillation its velocity is found to be $2m/s.$ Find the amplitude.

Answer 1

Hint: Amplitude is the maximum displacement which is related to the energy in the oscillation. When it is displaced from the equilibrium, the object performs the simple Harmonic motion and its maximum speed can be achieved when it passes through equilibrium. Use velocity to time relation equation for amplitude. Time period is the time taken by the body to complete one oscillation. It can be defined as the $T = \dfrac{{2\pi }}{\omega }$ where $\omega $ (omega) is the angular frequency and T is the time period. Find the correlation between the known values and unknown values asked.

Complete step by step answer:
Time period, $T = 16s$
$\omega = \dfrac{{2\pi }}{T}$
Place values and simplify –
$
\omega = \dfrac{{2\pi }}{{16}} \\
\Rightarrow\omega = \dfrac{\pi }{8}rad/s \\
 $
Velocity, $v = 2m/s$
At $t = 0$a particle passes through its mean position.
Now, for the maximum speed, the velocity to time relation can be given as-
$v = \omega A\cos \omega t$
Substitute the known values –
$ \Rightarrow 2 = \dfrac{\pi }{8}A\cos \left( {\dfrac{\pi }{8}} \right)\left( 2 \right)$
Simplify and make the unknown Amplitude, “A” the subject –
\[
2 = \dfrac{\pi }{8}A\cos \left( {45^\circ } \right)\left( 2 \right) \\
\Rightarrow 2 = \dfrac{\pi }{8}A\left( {\dfrac{1}{{\sqrt 2 }}} \right)\left( 2 \right) \\
\]
Do-cross-multiplication –
$
A = \dfrac{{2 \times 8 \times \sqrt 2 }}{\pi } \\
\Rightarrow A = \dfrac{{16\sqrt 2 }}{\pi } \\
$
Place the value of $\pi = 3.14$
$
A = \dfrac{{16\sqrt 2 }}{{3.14}} \\
\therefore A = 7.2m \\
$
Hence, the required answer – the Amplitude, $A = 7.2m$.

Note: Remember the basic trigonometric values of functions such as Sine and Cos and its different angles for direct substitutes. The above sum can be solved by different method as follows –
You can use equation –
$x = A\sin \left( {\dfrac{{2\pi }}{T}t} \right)$, where x is the distance,
Take derivatives on both sides with respect to “t”. As we know that distance per time is velocity and derivative of sin is cosine and again the derivative of angle with respect to t.
$v = A.\dfrac{{2\pi }}{T}\cos \left( {\dfrac{{2\pi }}{T}t} \right)$
Place all the given values and resultant will be the required answer.