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CBSE-XI-Physics

12: Simple Harmonic Motion

with Solutions - page 4

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  • Qstn #5
    The motion of a torsional pendulum is
    (a) periodic
    (b) oscillatory
    (c) simple harmonic
    (d) angular simple harmonic
    digAnsr:   a,b,d
    Ans : (a) periodic
    (b) oscillatory
    (d) angular simple harmonic
    Because it completes one oscillation in a fixed interval of time and the oscillations are in terms of rotation of the body through some angle.
    Page No 251:
  • Qstn #6
    Which of the following quantities are always negative in a simple harmonic motion?
    (a)
    F→. a→.
    (b)
    v→. r→.
    (c)
    a→. r→.
    (d)
    F→. r→.
    Ans : `` \left(c\right)\stackrel{\rightharpoonup }{a}.\stackrel{\rightharpoonup }{r}``
    `` \left(d\right)\stackrel{\rightharpoonup }{F}.\stackrel{\rightharpoonup }{r}``
    In S.H.M.,
    F = -kx
    Therefore, `` \stackrel{\rightharpoonup }{F.}\stackrel{\rightharpoonup }{r}`` will always be negative. As acceleration has the same direction as the force, `` \stackrel{\rightharpoonup }{a}.\stackrel{\rightharpoonup }{r}`` will also be negative, always.
    Page No 251:
  • Qstn #7
    Which of the following quantities are always positive in a simple harmonic motion?
    (a)
    F→. a→.
    (b)
    v→. r→.
    (c)
    a→. r→.
    (d)
    F→. r→.
    Ans : `` \left(a\right)\stackrel{\rightharpoonup }{F}.\stackrel{\rightharpoonup }{a}``
    `` ``
    As the direction of force and acceleration are always same, `` \stackrel{\rightharpoonup }{F}.\stackrel{\rightharpoonup }{a}``is always positive.
    Page No 251:
  • Qstn #8
    Which of the following quantities are always zero in a simple harmonic motion?
    (a)
    F→× a→.
    (b)
    v→× r→.
    (c)
    a→× r→.
    (d)
    F→× r→.
    Ans : `` \left(a\right)\stackrel{\rightharpoonup }{F}x\stackrel{\rightharpoonup }{a}``
    `` \left(b\right)\stackrel{\rightharpoonup }{v}x\stackrel{\rightharpoonup }{\stackrel{\rightharpoonup }{r}}``
    `` \left(c\right)\stackrel{\rightharpoonup }{a}x\stackrel{\rightharpoonup }{r}``
    `` \left(d\right)\stackrel{\rightharpoonup }{F}x\stackrel{\rightharpoonup }{r}``
    As `` \stackrel{\rightharpoonup }{F},\stackrel{\rightharpoonup }{a},\stackrel{\rightharpoonup }{r}and\stackrel{\rightharpoonup }{v}`` are either parallel or anti-parallel to each other, their cross products will always be zero.
    Page No 251:
  • Qstn #9
    Suppose a tunnel is dug along a diameter of the earth. A particle is dropped from a point, a distance h directly above the tunnel. The motion of the particle as seen from the earth is
    (a) simple harmonic
    (b) parabolic
    (c) on a straight line
    (d) periodic
    digAnsr:   c,d
    Ans : (c) on a straight line
    (d) periodic
    If the particle were dropped from the surface of the earth, the motion of the particle would be SHM. But when it is dropped from a height h, the motion of the particle is not SHM because there is no horizontal velocity imparted. In that case, the motion of the particle would be periodic and in a straight line.
    Page No 252:
  • Qstn #10
    For a particle executing simple harmonic motion, the acceleration is proportional to
    (a) displacement from the mean position
    (b) distance from the mean position
    (c) distance travelled since t = 0
    (d) speed
    digAnsr:   a
    Ans : (a) displacement from the mean position
    For S.H.M.,
    F = -kx
    ma = - kx (F = ma)
    or,
    a = `` -\frac{k}{m}x``
    Thus, acceleration is proportional to the displacement from the mean position but in opposite direction.
    Page No 252:
  • Qstn #11
    A particle moves in the X-Y plane according to the equation
    r→=i→+2j→Acosωt.The motion of the particle is
    (a) on a straight line
    (b) on an ellipse
    (c) periodic
    (d) simple harmonic
    digAnsr:   a,c,d
    Ans : (a) on a straight line
    (c) periodic
    (d) simple harmonic
    The given equation is a solution to the equation of simple harmonic motion. The amplitude is `` (\stackrel{\rightharpoonup }{i}+2\stackrel{\rightharpoonup }{j})A``, following equation of straight line y = mx + c. Also, a simple harmonic motion is periodic.
    Page No 252:
  • Qstn #12
    A particle moves on the X-axis according to the equation x = x0 sin2 ωt. The motion is simple harmonic
    (a) with amplitude x0
    (b) with amplitude 2x0
    (c) with time period
    2πω
    (d) with time period
    πω.
    digAnsr:   d
    Ans : (d) with time period `` \frac{\pi }{\omega }``
    Given equation:
    x = xo sin2 ωt
    ⇒​`` x=\frac{{x}_{0}}{2}(\,\mathrm{\,cos\,}2\omega t-1)``
    Now, the amplitude of the particle is xo/2 and the angular frequency of the SHM is 2ω.
    Thus, time period of the SHM = `` \frac{2\pi }{\text{angular frequency}}=\frac{2\pi }{2\omega }=\frac{\pi }{\omega }``
    Page No 252:
  • Qstn #13
    In a simple harmonic motion
    (a) the potential energy is always equal to the kinetic energy
    (b) the potential energy is never equal to the kinetic energy
    (c) the average potential energy in any time interval is equal to the average kinetic energy in that time interval
    (d) the average potential energy in one time period is equal to the average kinetic energy in this period.
    digAnsr:   d
    Ans : (d) the average potential energy in one time period is equal to the average kinetic energy in this period.
    The kinetic energy of the motion is given as,
    `` \frac{1}{2}k{A}^{2}{\,\mathrm{\,cos\,}}^{2}\omega t``
    The potential energy is calculated as,
    `` \frac{1}{2}k{A}^{2}{\,\mathrm{\,sin\,}}^{2}\omega t``
    As the average of the cosine and the sine function is equal to each other over the total time period of the functions, the average potential energy in one time period is equal to the average kinetic energy in this period.
    Page No 252:
  • Qstn #14
    In a simple harmonic motion
    (a) the maximum potential energy equals the maximum kinetic energy
    (b) the minimum potential energy equals the minimum kinetic energy
    (c) the minimum potential energy equals the maximum kinetic energy
    (d) the maximum potential energy equals the minimum kinetic energy
    digAnsr:   a,b
    Ans : (a) the maximum potential energy equals the maximum kinetic energy
    (b) the minimum potential energy equals the minimum kinetic energy
    In SHM,
    maximum kinetic energy = `` \frac{1}{2}k{A}^{2}``
    maximum potential energy = `` \frac{1}{2}k{A}^{2}``
    The minimum value of both kinetic and potential energy is zero.
    Therefore, in a simple harmonic motion the maximum kinetic energy and maximum potential energy are equal. Also, the minimum kinetic energy and the minimum potential energy are equal.
    Page No 252:
  • Qstn #15
    An object is released from rest. The time it takes to fall through a distance h and the speed of the object as it falls through this distance are measured with a pendulum clock. The entire apparatus is taken on the moon and the experiment is repeated
    (a) the measured times are same
    (b) the measured speeds are same
    (c) the actual times in the fall are equal
    (d) the actual speeds are equal
    digAnsr:   a,b
    Ans : (a) the measured times are same
    (b) the measured speeds are same
    The effect of gravity on the object as well as on the pendulum clock is same in both cases; the time measured is also same. As the time measured is same, the speeds are same.
    Page No 252:
  • Qstn #16
    Which of the following will change the time period as they are taken to moon?
    (a) A simple pendulum
    (b) A physical pendulum
    (c) A torsional pendulum
    (d) A spring-mass system
    digAnsr:   a,b
    Ans : (a) A simple pendulum
    (b) A physical pendulum
    As the time period of a simple pendulum and a physical pendulum depends on the acceleration due the gravity, the time period of these pendulums changes when they are taken to the moon.
    Page No 252:
  • #
    Section : iv
  • Qstn #1
    A particle executes simple harmonic motion with an amplitude of 10 cm and time period 6 s. At t = 0 it is at position x = 5 cm going towards positive x-direction. Write the equation for the displacement x at time t. Find the magnitude of the acceleration of the particle at t = 4 s.
    Ans : It is given,
    Amplitude of the simple harmonic motion, A =10 cm
    At t = 0 and x = 5 cm,
    Time period of the simple harmonic motion, T = 6 s
    Angular frequency (ω) is given by,
    `` \omega =\frac{2\,\mathrm{\,\pi \,}}{\,\mathrm{\,T\,}}=\frac{2\,\mathrm{\,\pi \,}}{6}=\frac{\,\mathrm{\,\pi \,}}{3}{\,\mathrm{\,sec\,}}^{-1}``
    Consider the equation of motion of S.H.M,
    Y = Asin `` \left(\omega t+\varphi \right)`` ...(1)
    where Y is displacement of the particle, and
    `` \varphi `` is phase of the particle.
    On substituting the values of A, t and ω in equation (1), we get:
    5 = 10sin(ω × 0 + ϕ)
    `` \Rightarrow ``5 = 10sin Ï•
    `` \,\mathrm{\,sin\,}\,\mathrm{\,\varphi \,}=\frac{1}{2}``
    `` \Rightarrow \,\mathrm{\,\varphi \,}=\frac{\,\mathrm{\,\pi \,}}{6}``
    ∴ Equation of displacement can be written as,
    `` x=\left(10\,\mathrm{\,cm\,}\right)\,\mathrm{\,sin\,}\left(\frac{\,\mathrm{\,\pi \,}}{3}t+\frac{\,\mathrm{\,\pi \,}}{6}\right)``
    (ii) At t = 4 s,
    `` x=10\,\mathrm{\,sin\,}\left[\frac{\,\mathrm{\,\pi \,}}{3}4+\frac{\,\mathrm{\,\pi \,}}{6}\right]``
    `` =10\,\mathrm{\,sin\,}\left[\frac{8\,\mathrm{\,\pi \,}+\,\mathrm{\,\pi \,}}{6}\right]``
    `` =10\,\mathrm{\,sin\,}\left(\frac{9\,\mathrm{\,\pi \,}}{6}\right)``
    `` =10\,\mathrm{\,sin\,}\left(\frac{3\,\mathrm{\,\pi \,}}{2}\right)``
    `` =10\,\mathrm{\,sin\,}\left(\pi +\frac{\,\mathrm{\,\pi \,}}{2}\right)``
    `` =-10\,\mathrm{\,sin\,}\frac{\,\mathrm{\,\pi \,}}{2}=-10``
    Acceleration is given by,
    a = -ω2x
    `` =\left(\frac{-{\,\mathrm{\,\pi \,}}^{2}}{9}\right)\times \left(-10\right)``
    `` =10.9\approx 11\,\mathrm{\,cm\,}/{\,\mathrm{\,sec\,}}^{-2}``
    Page No 252:
  • Qstn #2
    The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitude 2 cm, 1 m s-1 and 10 m s-2 at a certain instant. Find the amplitude and the time period of the motion.
    Ans : It is given that:
    Position of the particle, x = 2 cm = 0.02 m
    Velocity of the particle, v = 1 ms-1.
    Acceleration of the particle, a = 10 ms-2.
    Let `` \omega `` be the angular frequency of the particle.
    The acceleration of the particle is given by,
    a = ω2x
    `` \Rightarrow \omega =\sqrt{\frac{a}{x}}=\sqrt{\frac{10}{0.02}}``
    `` =\sqrt{500}=10\sqrt{5}\,\mathrm{\,Hz\,}``
    `` \,\mathrm{\,Time\,}\,\mathrm{\,period\,}\,\mathrm{\,of\,}\,\mathrm{\,the\,}\,\mathrm{\,motion\,}\,\mathrm{\,is\,}\,\mathrm{\,given\,}\,\mathrm{\,as\,},``
    `` T=\frac{2\,\mathrm{\,\pi \,}}{\,\mathrm{\,\omega \,}}=\frac{2\,\mathrm{\,\pi \,}}{10\sqrt{5}}``
    `` =\frac{2\times 3.14}{10\times 2.236}``
    `` =0.28\,\mathrm{\,s\,}``
    Now, the amplitude A is calculated as,
    `` v=\omega \sqrt{{A}^{2}-{x}^{2}}``
    `` ``
    `` \Rightarrow {v}^{2}={\omega }^{2}\left({A}^{2}-{x}^{2}\right)``
    `` 1=500\left({A}^{2}-0.0004\right)``
    `` \Rightarrow A=0.0489=0.049\,\mathrm{\,m\,}``
    `` \Rightarrow A=4.9\,\mathrm{\,cm\,}``
    Page No 252:
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