Question

An air column, closed at one end and open at the other, resonates with a tunning fork when the smallest length of the column is $$50\,cm.$$  The next larger length of the column resonating with the same tunning fork is

A. $$100\,cm$$
B. $$150\,cm$$  
C. $$200\,cm$$
D. $$66.7\,cm$$
Answer :   $$150\,cm$$
Solution :
The smallest length of the air column is associated with fundamental mode of vibration of the air column as shown in the diagram.
Simple Harmonic Motion (SHM) mcq solution image
$$\because {L_{\min }} = \frac{\lambda }{4} \Rightarrow 50\,cm = \frac{\lambda }{4} \Rightarrow \lambda = 200\,cm$$
The next higher length of the air column is
$$\eqalign{ & L = \frac{\lambda }{4} + \frac{\lambda }{2} = \frac{{\lambda + 2\lambda }}{4} = \frac{{3\lambda }}{4} \cr & = \frac{3}{4} \times 200 = 150\,cm \cr} $$

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. $${T_1} = 2{T_2}$$
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Simple Harmonic Motion (SHM)

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