Question 166: Oscillations

Question

A particle performs simple harmonic motion with amplitude $$A$$. Its speed is increased to three times at an instant when its displacement is $$\frac{2 A}{3}$$. The new amplitude of motion is $$\frac{n A}{3}$$. The value of $$n$$ is ___________.

Accepted Answer

To find the new amplitude of the motion when the speed is increased to three times at a given displacement, we use the concepts of simple harmonic motion (SHM) and its formulas.
In SHM, the velocity $v$ of a particle at a displacement $x$ from the mean position can be given by the formula:
$$v = \omega \sqrt{A^2 - x^2}$$
where:
$\omega$ is the angular frequency of the motion,
$A$ is the amplitude, and
$x$ is the displacement at that instance.
Given:
Displacement at the instance, $x = \frac{2A}{3}$,
Initial velocity is increased to three times at this displacement.
Thus, let's find the initial velocity $v$ at $x = \frac{2A}{3}$:
$$v = \omega \sqrt{A^2 - \left(\frac{2A}{3}ight)^2} = \omega \sqrt{A^2 - \frac{4A^2}{9}} = \omega \sqrt{\frac{5A^2}{9}} = \frac{\sqrt{5}A\omega}{3}$$
With the velocity increased to three times, the new velocity $v'$ becomes:
$$v' = 3v = 3 	imes \frac{\sqrt{5}A\omega}{3} = \sqrt{5}A\omega$$
For the new amplitude $A'$, the velocity $v'$ at the same displacement $x$ is:
$$v' = \omega \sqrt{{A'}^2 - \left(\frac{2A}{3}ight)^2}$$
Setting the expressions for $v'$ equal gives:
$$\sqrt{5}A\omega = \omega \sqrt{{A'}^2 - \frac{4A^2}{9}}$$
$$\sqrt{5}A = \sqrt{{A'}^2 - \frac{4A^2}{9}}$$
Solving for $A'$ gives:
$${A'}^2 = 5A^2 + \frac{4A^2}{9} = \frac{45A^2 + 4A^2}{9} = \frac{49A^2}{9}$$
$$A' = \sqrt{\frac{49A^2}{9}} = \frac{7A}{3}$$
Therefore, the new amplitude of the motion is $\frac{7A}{3}$, which means the value of $n$ is 7.

Oscillations

A particle performs simple harmonic motion with amplitude $A. Its speed is increased to three times at an instant when its displacement is \frac{2 A}{3}. The new amplitude of motion is \frac{n A}{3}. The value of n$ is ___________.

Integer / Numeric Answer Type