1 Marks Question

Question 11 Mark

The energy of a system in simple harmonic motion is given by $\text{E}=\frac{1}{2}\text{m }\omega^2\text{A}^2.$ Which of the following two statements is more appropriate?

The energy is increased because the amplitude is increased.
The amplitude is increased because the energy is increased.

Answer

Statement A is more appropriate because the energy of a system in simple harmonic motion is given by $\text{E}=\frac{1}{2}\text{m }\omega^2\text{A}^2.$
If the mass (m) and angular frequency $(\omega)$ are made constant, Energy (E) becomes proportional to the square of amplitude (A²).
i.e. E $\propto$ A²
Therefore, according to the relation, energy increases as the amplitude increases.

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Question 21 Mark

A platoon of soldiers marches on a road in steps according to the sound of a marching band. The band is stopped and the soldiers are ordered to break the steps while crossing a bridge. Why?

Answer

Forced oscillation may break the bridge.

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Question 31 Mark

A pendulum clock gives correct time at the equator. Will it gain time or loose time as it is taken to the poles?

Answer

$\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{g}}}$ at pole g is more so time period gets decreased hence check gains time.

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Question 41 Mark

A particle executing simple harmonic motion comes to rest at the extreme positions. Is the resultant force on the particle zero at these positions according to Newton's first law?

Answer

No, force is maximum at extreme position.

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Question 51 Mark

Can simple harmonic motion take place in a noninertial frame? If yes, should the ratio of the force applied with the displacement be constant?

Answer

Yes. Simple harmonic motion can take place in a non-inertial frame. However, the ratio of the force applied to the displacement cannot be constant because a non-inertial frame has some acceleration with respect to the inertial frame. Therefore, a fictitious force should be added to explain the motion.

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Question 61 Mark

Can the potential energy in a simple harmonic motion be negative? Will it be so if we choose zero potential energy at some point other than the mean position?

Answer

No. It cannot be negative because the minimum potential energy of a particle executing simple harmonic motion at mean position is zero. The potential energy increases in positive direction at the extreme position.
However, if we choose zero potential energy at some other point, say extreme position, the potential energy can be negative at the mean position.

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Question 71 Mark

A small creature moves with constant speed in a vertical circle on a bright day. Does its shadow formed by the sun on a horizontal plane move in a simple harmonic motion?

Answer

Yes, its shadow on a horizontal plane moves in simple harmonic motion. The projection of a uniform circular motion executes simple harmonic motion along its diameter (which is the shadow on the horizontal plane), with the mean position lying at the centre of the circle.

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Question 81 Mark

A particle executes simple harmonic motion. Let P be a point near the mean position and Q be a point near an extreme. The speed of the particle at P is larger than the speed at Q. Still the particle crosses P and Q equal number of times in a given time interval. Does it make you unhappy?

Answer

No. It does not make me unhappy because the number of times a particle crosses the mean and extreme positions does not depend on the speed of the particle.

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Question 91 Mark

Can a pendulum clock be used in an earth satellite?

Answer

No. According to the relation:

$\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{g}}}$

The time period of the pendulum clock depends upon the acceleration due to gravity. As the earth-satellite is a free falling body and its g (effective acceleration due to gravity) is zero at the satellite, the time period of the clock is infinite.

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Question 101 Mark

A block of known mass is suspended from a fixed support through a light spring. Can you find the time period of vertical oscillation only by measuring the extension of the spring, when the block is in equilibrium?

Answer

Yes. $\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{k}}}=2\pi\sqrt{\frac{\text{x}_0}{\text{g}}}\text{as gm}=\text{kx}_0.$

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Question 111 Mark

The springs shown in the figure are all unstretched in the beginning when a man starts pulling the block. The man exerts a constant force F on the block. Find the amplitude and the frequency of the motion of the block.

Answer

K₂and K₃ are in series.

Let equivalent spring constant be K₄

$\therefore\frac{1}{\text{K}_4}=\frac{1}{\text{K}_2}+\frac{1}{\text{K}_3}=\frac{\text{K}_2+\text{K}_3}{\text{K}_2\text{K}_3}$

$\Rightarrow\text{K}_4=\frac{\text{K}_2\text{K}_3}{\text{K}_2+\text{K}_3}$

Now K₄ and K₁ are in parallel.

So equivalent spring constant $\text{K}=\text{K}_1+\text{K}_4=\frac{\text{K}_2\text{K}_3}{\text{K}_2+\text{K}_3}+\text{K}_1$

$=\frac{\text{K}_2\text{K}_3+\text{K}_1\text{K}_2+\text{K}_1\text{K}_3}{\text{K}_2+\text{K}_3}$

$\therefore\text{T}=2\pi\sqrt{\frac{\text{M}}{\text{k}}}=2\pi\sqrt{\frac{\text{M(k}_2+\text{k}_3)}{\text{k}_2\text{k}_3+\text{k}_1\text{k}_2+\text{k}_1\text{k}_3}}$

Frequency $=\frac{1}{\text{T}}=\frac{1}{2\pi}\sqrt{\frac{\text{k}_2\text{k}_3+\text{k}_1\text{k}_2+\text{k}_1\text{k}_3}{\text{M}\text{(k}_2+\text{k}_3)}}$
Amplitube $\text{x}=\frac{\text{F}}{\text{k}}=\frac{\text{F}(\text{k}_2+\text{k}_3)}{\text{k}_1\text{k}_2+\text{k}_2\text{k}_3+\text{k}_1\text{k}_3}$

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