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Simple Harmonic Motion question

2024 · 31 Jan · Shift 1 · Q86

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Simple Harmonic Motion
2024 · 31 Jan · Shift 1 · Q86

Simple Harmonic Motion question

2024 · 31 Jan · Shift 1 · Q86

JEE MainPhysicsSimple Harmonic MotionNumerical+4 / −1

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

\underline{\hspace{2cm}}

.

Numerical answer

View written solutionFree

Correct answer: 7

Initial SHM relation

For a particle in SHM with amplitude $A$ and angular frequency $\omega$ , the speed at displacement $x$ is

$v=\omega\sqrt{A^2-x^2}.$

At the given instant,

$x=\frac{2A}{3}.$

So the original speed is

=\omega\sqrt{A^2-\frac{4A^2}{9}} =\omega\sqrt{\frac{5A^2}{9}} =\frac{\omega A\sqrt{5}}{3}.$$ 2. **Speed is made three times** The new speed becomes $$v'=3v_0=3\cdot \frac{\omega A\sqrt{5}}{3}=\omega A\sqrt{5}.$$ 3. **Find the new amplitude** Immediately after increasing the speed, the displacement is still $$x=\frac{2A}{3},$$ but the motion now has a new amplitude, say $A'$. For SHM, $$v'^2=\omega^2\left(A'^2-x^2\right).$$ Substitute the values: $$\left(\omega A\sqrt{5}\right)^2=\omega^2\left(A'^2-\left(\frac{2A}{3}\right)^2\right).$$ $$5\omega^2A^2=\omega^2\left(A'^2-\frac{4A^2}{9}\right).$$ Cancel $\omega^2$: $$5A^2=A'^2-\frac{4A^2}{9}.$$ So, $$A'^2=5A^2+\frac{4A^2}{9}=\frac{45A^2+4A^2}{9}=\frac{49A^2}{9}.$$ Hence, $$A'=\frac{7A}{3}.$$ 4. **Compare with given form** Given new amplitude is $$\frac{nA}{3},$$ so $$\frac{nA}{3}=\frac{7A}{3} \implies n=7.$$ ## Final Answer $$\boxed{7}$$ The derived answer matches the stored correct answer.

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