A body hanging from a spring can oscillate in a horizontal plane … - Vidyadip

Question

A body hanging from a spring can oscillate in a horizontal plane with angular velocity $\omega$ without friction or damping when it is stretched to a distance and then released. It passes through the equilibrium center with a velocity at time $t=0$. Find the amplitude of the resultant oscillation in terms of parameter $\omega_0, x_0$ and $v_0$.
[Hint : The equation $x=a \cos (\omega t+\theta)$ is negative keep in mind that the initial]

✓

Answer

Let at any time t
$x=A \cos (\omega t+\theta)$
$A =$ Amplitude, $\theta=$ initial phase angle
Velocity $\frac{d x}{d t}=- A \omega \sin (\omega t +\theta)$
when $t=0, x=x_0$ and $\frac{d x}{d t}=- v _0$
$\begin{aligned}x_0 & =A \cos \theta\quad\ldots\ldots (1) \\-v_0 & =-A \omega \sin \theta \\A \sin \theta & =\frac{v_0}{\omega}\quad\ldots\ldots (2)\end{aligned}$
By squaring equation (1) and (2) and adding
$\begin{array}{l}( A \cos \theta)^2+( A \sin \theta)^2=x_0^2+\left(\frac{ v _0}{\omega}\right)^2 \\ A^2\left(\cos ^2 \theta+\sin ^2 \theta\right)=x_0^2+\frac{ v _0^2}{\omega^2} \\ A^2 \times 1=x_0^2+\frac{ v _0^2}{\omega^2} \\ A=\left(x_0^2+\frac{ v _0^2}{\omega^2}\right)^{\frac{1}{2}}\end{array}$

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