Question

A pendulum with time period of $$1s$$  is losing energy. At certain time its energy is $$45\,J.$$  If after completing 15 oscillations, its energy has become $$15\,J,$$  its damping constant (in $${s^{ - 1}}$$) is:

A. $$\frac{1}{2}$$
B. $$\frac{1}{{30}}\ln 3$$
C. 2
D. $$\frac{1}{{15}}\ln 3$$  
Answer :   $$\frac{1}{{15}}\ln 3$$
Solution :
$$\eqalign{ & {\text{As we know,}}\,E = {E_0}{e^{ - \frac{{bt}}{m}}} \cr & 15 = 45{e^{ - \frac{{b15}}{m}}} \Rightarrow \frac{1}{3} = {e^{ - \frac{{b15}}{m}}}\,\,\left[ {{\text{As no}}{\text{. of oscillations}} = 15{\text{ so}}\,t = 15\,\sec } \right] \cr} $$
Taking log on both sides
$$\frac{b}{m} = \frac{1}{{15}}\ln 3$$

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. $$\sqrt {\frac{{{k_2}}}{{{k_1}}}} $$
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
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. $$2\pi \sqrt {\frac{L}{{g\tan \alpha }}} $$
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. $${T_1} = 2{T_2}$$

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


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