Question

What will be the force constant of the spring system shown in figure?
Simple Harmonic Motion (SHM) mcq question image

A. $$\frac{{{k_1}}}{2} + {k_2}$$
B. $${\left[ {\frac{1}{{2{k_1}}} + \frac{1}{{{k_2}}}} \right]^{ - 1}}$$  
C. $$\left[ {\frac{1}{{2{k_1}}} + \frac{1}{{{k_2}}}} \right]$$
D. $${\left[ {\frac{2}{{{k_1}}} + \frac{1}{{{k_2}}}} \right]^{ - 1}}$$
Answer :   $${\left[ {\frac{1}{{2{k_1}}} + \frac{1}{{{k_2}}}} \right]^{ - 1}}$$
Solution :
Two springs of force constants $${{k_1}}$$ and $${{k_1}}$$ are in parallel. Hence
$$k' = {k_1} + {k_1} = 2\,{k_1}$$
The third spring $${k_2}$$ is in series with spring of force constant $${k'}.$$
$$\therefore \frac{1}{k} = \left[ {\frac{1}{{2{k_1}}} + \frac{1}{{{k_2}}}} \right]\,\,{\text{or}}\,\,k = {\left[ {\frac{1}{{2{k_1}}} + \frac{1}{{{k_2}}}} \right]^{ - 1}}$$

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

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Simple Harmonic Motion (SHM)

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