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Simple Harmonic Motion question

2016 · Shift 0 · Q54

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JEE Main
Physics
Simple Harmonic Motion
2016 · Shift 0 · Q54

Simple Harmonic Motion question

2016 · Shift 0 · Q54

JEE MainPhysicsSimple Harmonic MotionMCQ+4 / −1

A particle performs simple harmonic motion with amplitude

A.

Its speed is trebled at the instant that it is at a distance

{{2A} \over 3}

from equilibrium position. The new amplitude of the motion is:

A
$A\sqrt 3$
B
${{7A} \over 3}$
C
${A \over 3}\sqrt {41}$
D
$3A$

View written solutionFree

Correct answer: B

Use the SHM speed formula

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

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

Here,

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

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}.$$ 2. **Speed is trebled** The new speed becomes $$v' = 3v = 3\cdot \frac{\omega A\sqrt{5}}{3} = \omega A\sqrt{5}.$$ 3. **Find the new amplitude using energy relation** For SHM, if at displacement $x$ the speed is $v'$, then the new amplitude $A'$ satisfies $$v'^2 = \omega^2\left(A'^2 - x^2\right).$$ Substitute the known values: $$\omega^2 A^2 \cdot 5 = \omega^2\left(A'^2 - \left(\frac{2A}{3}\right)^2\right).$$ Cancel $\omega^2$: $$5A^2 = A'^2 - \frac{4A^2}{9}.$$ Hence, $$A'^2 = 5A^2 + \frac{4A^2}{9} = \frac{45A^2 + 4A^2}{9} = \frac{49A^2}{9}.$$ Therefore, $$A' = \frac{7A}{3}.$$ 4. **Check options** - A: $A\sqrt{3}$ - B: $\dfrac{7A}{3}$ - C: $\dfrac{A\sqrt{41}}{3}$ - D: $3A$ The correct option is $$\boxed{\frac{7A}{3}}.$$ So **Option B** is correct.

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