a pendulum is displaced to an angle theta from its

A pendulum is displaced to an angle $$\theta $$ from its equilibrium position, then it will pass through its mean position with a velocity $$v$$ equal to

A. $$\sqrt {2gl} $$

B. $$\sqrt {2gl\sin \theta } $$

C. $$\sqrt {2gl\cos \theta } $$

D. $$\sqrt {2gl\left( {1 - \cos \theta } \right)} $$

Answer : $$\sqrt {2gl\left( {1 - \cos \theta } \right)} $$

Solution :
If $$l$$ is the length of pendulum and $$\theta $$ the angular amplitude, then height
Simple Harmonic Motion (SHM) mcq solution image

$$\eqalign{ & h = AB - AC \cr & = l - l\cos \theta \cr & = l\left( {1 - \cos \theta } \right)\,......\left( {\text{i}} \right) \cr} $$
At point $$P$$ (maximum displacement position i.e. extreme position), potential energy is maximum and kinetic energy is zero. At point $$B$$ (mean or equilibrium position) potential energy is minimum and kinetic energy is maximum, so from principle of conservation of energy.
$$\eqalign{ & \left( {PE + KE} \right)\,{\text{at}}\,P = \left( {KE + PE} \right)\,{\text{at}}\,B \cr & {\text{or}}\,\,mgh + 0 = \frac{1}{2}m{v^2} + 0 \cr & {\text{or}}\,\,v = \sqrt {2gh} \,......\left( {{\text{ii}}} \right) \cr} $$
Substituting the value of $$h$$ from Eq. (i) into Eq. (ii),
$$v = \sqrt {2gl\left( {1 - \cos \theta } \right)} $$

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

A pendulum is displaced to an angle $$\theta $$ from its equilibrium position, then it will pass through its mean position with a velocity $$v$$ equal to

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

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

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 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

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

Practice More MCQ Question on Physics Section

A pendulum is displaced to an angle $$\theta $$ from its equilibrium position, then it will pass through its mean position with a velocity $$v$$ equal to

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

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

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 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

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

Practice More MCQ Question on Physics Section

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

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