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CBSE-XI-Physics

12: Simple Harmonic Motion

with Solutions -

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    Section : i
  • Qstn #1
    A person goes to bed at sharp 10.00 pm every day. Is it an example of periodic motion? If yes, what is the time period? If no, why?
    Ans : No. As motion is a change in position of an object with respect to time or a reference point, it is not an example of periodic motion.
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  • Qstn #2
    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?
    Ans : No. The resultant force on the particle is maximum at the extreme positions.
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  • Qstn #3
    Can simple harmonic motion take place in a non-inertial frame? If yes, should the ratio of the force applied with the displacement be constant?
    Ans : 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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  • Qstn #4
    A particle executes simple harmonic motion. If you are told that its velocity at this instant is zero, can you say what is its displacement? If you are told that its velocity at this instant is maximum, can you say what is its displacement?
    Ans : No, we cannot say anything from the given information. To determine the displacement of the particle using its velocity at any instant, its mean position has to be known.
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  • Qstn #5
    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 sample harmonic motion?
    Ans : 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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  • Qstn #6
    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?
    Ans : 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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  • Qstn #7
    In measuring time period of a pendulum, it is advised to measure the time between consecutive passage through the mean position in the same direction. This is said to result in better accuracy than measuring time between consecutive passage through an extreme position. Explain.
    Ans : The mean position of a particle executing simple harmonic motion is fixed, whereas its extreme position keeps on changing. Therefore, when we use stop watch to measure the time between consecutive passage, we are certain about the mean position.
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  • Qstn #8
    It is proposed to move a particle in simple harmonic motion on a rough horizontal surface by applying an external force along the line of motion. Sketch the graph of the applied force against the position of the particle. Note that the applied force has two values for a given position depending on whether the particle is moving in positive or negative direction.
    Ans : Figure
    (a) shows the graph of the applied force against the position of the particle.
    (a)
    (b)
    (c)
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  • Qstn #9
    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?
    Ans : 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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  • Qstn #10
    The energy of system in simple harmonic motion is given by
    E=12m ω2A2.Which of the following two statements is more appropriate?
    (A) The energy is increased because the amplitude is increased.
    (B) The amplitude is increased because the energy is increased.
    Ans : Statement A is more appropriate because the energy of a system in simple harmonic motion is given by `` E=\frac{1}{2}m{\omega }^{2}{A}^{2}.``
    If the mass (m) and angular frequency (ω) are made constant, Energy (E) becomes proportional to the square of amplitude (A2).
    i.e. E ∝ A2
    Therefore, according to the relation, energy increases as the amplitude increases.
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  • Qstn #11
    A pendulum clock gives correct time at the equator. Will it gain time or loose time as it is taken to the poles?
    Ans : According to the relation:
    `` T=2\pi \sqrt{\frac{l}{g}}``
    The time period (T) of the pendulum becomes proportional to the square root of inverse of g if the length of the pendulum is kept constant.
    i.e. `` T\propto \sqrt{\frac{1}{g}}``
    Also, acceleration due to gravity (g) at the poles is more than that at equator. Therefore, the time period decreases and the clock gains time.
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  • Qstn #12
    Can a pendulum clock be used in an earth-satellite?
    Ans : No. According to the relation:
    `` T=2\pi \sqrt{\frac{l}{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 geffective (effective acceleration due to gravity ) is zero at the satellite, the time period of the clock is infinite.
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  • Qstn #13
    A hollow sphere filled with water is used as the bob of a pendulum. Assume that the equation for simple pendulum is valid with the distance between the point of suspension and centre of mass of the bob acting as the effective length of the pendulum. If water slowly leaks out of the bob, how will the time period vary?
    Ans : The time period of a pendulum depends on the length and is given by the relation, `` T=2\pi \sqrt{\frac{l}{g}}``.
    As the effective length of the pendulum first increases and then decreases, the time period first increases and then decreases.
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