Question

A particle moving with simple harmonic motion has speed $3cm/s$ and $4cm/s$ at displacement $8cm$ and $6cm$, respectively, from the equilibrium position. Find
(a) the period of oscillation, and
(b) the amplitude of oscillation.

Answer 1

Using the relation

$v=w\sqrt{A^2-x^2}$ [$v$ is the velocity as displacement $x$, $A$ :Amplitude]

From given

(i) when $x=8,v=3$

$3=w\sqrt{A^2-64}$ ........... $\left(1\right)$

(ii) when $x=6,v=4$

$4=w\sqrt{A^2-36}$ .......... $\left(2\right)$

dividing $\left(1\right)÷\left(2\right)$ gives

${\left(\frac{3}{4}\right)}^2=\frac{A^2-64}{A^2-36}\Rightarrow A^2=\frac{16\times 64-9\times 36}{7}=100$

$\Rightarrow A=10cm$

Putting $A$ in $\left(1\right)$ gives

$3=w\sqrt{{10}^2-64}\Rightarrow w=\frac{3}{6}=\frac{1}{2}rad/s$

$\therefore$ Time period ocsillaion

$T=\frac{2\pi }{w}=4\pi$