Question

A particle performs simple harmonic mition with amplitude $$A.$$ Its speed is trebled at the instant that it is at a distance $$\frac{{2A}}{3}$$ from equilibrium position. The new amplitude of the motion is

A. $$A\sqrt 3 $$
B. $$\frac{{7A}}{3}$$  
C. $$\frac{A}{3}\sqrt {41} $$
D. $$3A$$
Answer :   $$\frac{{7A}}{3}$$
Solution :
We know that $$V = \root \omega \of {{A^2} - {x^2}} $$
Initially $$V = \root \omega \of {{A^2} - {{\left( {\frac{{2A}}{3}} \right)}^2}} $$
Finally $$3v = \root \omega \of {A{'^2} - {{\left( {\frac{{2A}}{3}} \right)}^2}} $$
Where $$A' = $$ final amplitude (Given at $$x = \frac{{2A}}{3},$$   velocity to trebled)
On dividing we get $$\frac{3}{1} = \frac{{\sqrt {A{'^2} - {{\left( {\frac{{2A}}{3}} \right)}^2}} }}{{\sqrt {{A^2} - {{\left( {\frac{{2A}}{3}} \right)}^2}} }}$$
$$9\left[ {{A^2} - \frac{{4{A^2}}}{9}} \right] = A{'^2} - \frac{{4{A^2}}}{9}\,\,\,\,\,\,\,\,\,\,\,\therefore A' = \frac{{7A}}{3}$$

Releted MCQ Question on
Oscillation and Mechanical Waves >> Simple Harmonic Motion (SHM)

Releted Question 1

Two bodies $$M$$ and $$N$$ of equal masses are suspended from two separate massless springs of spring constants $${k_1}$$ and $${k_2}$$ respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of vibration of $$M$$ to that of $$N$$ is

A. $$\frac{{{k_1}}}{{{k_2}}}$$
B. $$\sqrt {\frac{{{k_1}}}{{{k_2}}}} $$
C. $$\frac{{{k_2}}}{{{k_1}}}$$
D. $$\sqrt {\frac{{{k_2}}}{{{k_1}}}} $$
View Answer
Releted Question 2

A particle free to move along the $$x$$-axis has potential energy given by $$U\left( x \right) = k\left[ {1 - \exp \left( { - {x^2}} \right)} \right]$$      for $$ - \infty \leqslant x \leqslant + \infty ,$$    where $$k$$ is a positive constant of appropriate dimensions. Then

A. at points away from the origin, the particle is in unstable equilibrium
B. for any finite nonzero value of $$x,$$ there is a force directed away from the origin
C. if its total mechanical energy is $$\frac{k}{2},$$  it has its minimum kinetic energy at the origin.
D. for small displacements from $$x = 0,$$  the motion is simple harmonic
View Answer
Releted Question 3

The period of oscillation of a simple pendulum of length $$L$$ suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination $$\alpha ,$$ is given by

A. $$2\pi \sqrt {\frac{L}{{g\cos \alpha }}} $$
B. $$2\pi \sqrt {\frac{L}{{g\sin \alpha }}} $$
C. $$2\pi \sqrt {\frac{L}{g}} $$
D. $$2\pi \sqrt {\frac{L}{{g\tan \alpha }}} $$
View Answer
Releted Question 4

A particle executes simple harmonic motion between $$x = - A$$  and $$x = + A.$$  The time taken for it to go from 0 to $$\frac{A}{2}$$ is $${T_1}$$ and to go from $$\frac{A}{2}$$ to $$A$$ is $${T_2.}$$ Then

A. $${T_1} < {T_2}$$
B. $${T_1} > {T_2}$$
C. $${T_1} = {T_2}$$
D. $${T_1} = 2{T_2}$$
View Answer

Practice More Releted MCQ Question on
Simple Harmonic Motion (SHM)

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Basic PhysicsUnit and MeasurementKinematicsLaws of MotionFrictionUniform Circular MotionImpulseMomentumWork Energy and PowerRotational MotionGravitationMechanical Properties of Solids and Fluids
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Oscillation and Mechanical WavesSimple Harmonic Motion (SHM)Waves