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Waves Test - 75

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Waves Test - 75
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  • Question 1
    1 / -0
    Maximum acceleration of an object in simple harmonic motion is 24 m/s2m/s^2 and maximum velocity is 16 m/sec. The amplitude of object is 
    Solution

  • Question 2
    1 / -0

    Directions For Questions

    A block of mass MM is suspended from on end of a light spring as shown. The origin OO is considered at distance equal to the natural length of the spring from the ceiling and vertical downward direction as positive yy- axis. When the system is in equilibrium, a bullet of mass m/3m/3 moving in a vertical upward direction with velocity v0v_{0} strikes the block and embeds into it. As a result, the block (with a bullet embedded into it) moves up and starts oscillating.
    Based on the given information answer the following question:

    ...view full instructions

    The time taken by block-bullet system to move from y=mgky = \dfrac { m g } { k } (initial equilibrium position) to y=0y = 0 (natural length of spring) is (AA represents the amplitude of motion).
    Solution
    Initially in equilibrium let the elongation is spring be y0y_{0} then mg=ky0mg=ky_{0}
    y0=mgky_{0}=\dfrac{mg}{k}
    As the bullet strikes the block with velocity v0v_{0} and gets embedded into it, the velocity of the combined mass can be computed by using the principle of moment conservation.
    m3v0=4m3vv=v04\dfrac{m}{3}v_{0}=\dfrac{4m}{3}v\Rightarrow v=\dfrac{v_{0}}{4}
    Let new mean position is at distance yy from the origin, then
    ky=4m3gy=4mg3kky=\dfrac{4m}{3}g\Rightarrow y=\dfrac{4 mg}{3k}

    Now , the block executes SHMSHM about mean position defined by y=4mg/3ky=4 mg/3k with time period T=2π4m/3kT=2\pi \sqrt{4m/3k}. At t=0t=0, the combined mass is at a displacement of (yy0)(y-y_{0}) from mean position and is moving with velocity vv, then by using v=ωA2x2v=\omega \sqrt{A^{2}-x^{2}}, we can find the amplitude of motion.

    (v04)2=3k4m[A2(yy0)2]=3k4m[A2(mg3k)2]\left(\dfrac{v_{0}}{4}\right)^{2}=\dfrac{3k}{4m}[A^{2}-(y-y_{0})^{2}]=\dfrac{3k}{4m}\left[A^{2}-\left(\dfrac{mg}{3k}\right)^{2}\right]

    A=mv0212k+(mg3k)2\Rightarrow A=\sqrt{\dfrac{mv_{0}^{2}}{12k}+\left(\dfrac{mg}{3k}\right)^{2}}
    To compute the time taken by the combined mass from y=mg/ky=mg/k to y=0y=0, we can either go for equation method or circuit motion projection method.

    Required time, t=θω=αβωt=\dfrac{\theta}{\omega}=\dfrac{\alpha-\beta}{\omega}

    cosa=yy0A=(4mg3k4mgkA)=mg3kA\cos a=\dfrac{y-y_{0}}{A}=\left(\dfrac{\dfrac{4mg}{3k}-\dfrac{4mg}{k}}{A}\right)=\dfrac{mg}{3kA}

    cosβ=yA=4mg3kA\cos\beta=\dfrac{y}{A}=\dfrac{4mg}{3kA}

    So, t=cos(mg3kA)cos1(4mg3kA)ωt=\dfrac{\cos\left(\dfrac{mg}{3kA}\right)-\cos^{-1}\left(\dfrac{4mg}{3kA}\right)}{\omega}
          t=4m3k[cos1(mg3kA)cos1(4mg3kA)]t=\sqrt{\dfrac{4m}{3k}}\left[\cos^{-1}\left(\dfrac{mg}{3kA}\right)-\cos^{-1}\left(\dfrac{4mg}{3kA}\right)\right]

  • Question 3
    1 / -0
    A person observe two points on a string as a travelling wave passes them. The points are at x1=0x _ { 1 } = 0 and x2=1mx_2 = 1m. The transverse motions of the two points are found to be as follows: y1=0.2sin3πty _ { 1 } = 0.2 \sin 3 \pi t
    y2=0.2sin(3πt+π/8)y _ { 2 } = 0.2 \sin ( 3 \pi t + \pi/8 ) What is the frequency in HzHz?
    Solution

  • Question 4
    1 / -0
    A wave pulse is given by the equation y=f(x,t)=Aexp(B(xvt)2)y=f(x,t)=A exp(-B(x-vt)^{2}). Given A=1.0m.B=1.0m2A=1.0m.B=1.0m^{-2} and v=+2.0m/sv=+2.0m/s. which of the following graph shows the correct wave profile at the instant t=1st=1s?
    Solution

  • Question 5
    1 / -0
    A particle starts from the origin, goes along XX- axis to the point (20 m, 0)(20\ m,\ 0) and then returns along the same line to the point (20m, 0)(-20m,\ 0). The distance and displacement of the particle during the trip are
  • Question 6
    1 / -0
    A particle executes simple harmonic motion with a time period of 16 s. At time t=2s, the particle crosses the mean position. Its velocity is 4ms14 ms^{-1} when t=4s. The amplitude of motion is 
    Solution

  • Question 7
    1 / -0
    A particle executes SHM with a time period of 16 second .at time t is equal 2 second the particle crosses the mean position while at t is equal to 4 second its velocity is 4 m per second the amplitude of motion in metre is
    Solution

  • Question 8
    1 / -0
    The distance between consecutive maxima and minima is given by
  • Question 9
    1 / -0
    The amplitude of a wave represented by displacement equation :
     y=1a sinωt±1b cosωt y=\frac { 1 }{ \sqrt { a }  } sin\omega t\pm \frac { 1 }{ \sqrt { b }  } cos\omega t will be :
    Solution

  • Question 10
    1 / -0
    A body is performing linear SHM, If the displacement, acceleration and the corresponding velocity of the body are y, 'a' and v respectievly, which of the following graphs is/are correct?
    Solution

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