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Waves Test - 61

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Waves Test - 61
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  • Question 1
    1 / -0
    A particle performs simple harmonic motionwith amplitude A. Its speed is trebled at the instant when it is at a distance 2A3\dfrac { 2 \mathrm { A } } { 3 } fromequilibrium position. Find the new amplitudeof the motion. 
  • Question 2
    1 / -0
    A sound wave travels at a speed of 330m/s.330m/s. If the wavelength is 1.56cm1.56cm  what is the frequency of the wave? will it be audible for human
    Solution

    s=330s=330

    h=1.5cm=1.5/100=0.015h=1.5cm=1.5/100=0.015

    s×hs×h

    330×0.015330×0.015

    =4.95Hz=4.95 Hz

    it Will not be audible as it less than 20Hz20Hz

    Hence,

    option (C)(C) is correct answer.


  • Question 3
    1 / -0
    A particle in S.H.M. is described by the displacement function x(t)=acos(ax+θ)x(t) = a cos(ax + \theta). If the initial (t = 0) position of the particle is 1 cm and its initial velocity is πcm/s\pi cm/s. The angular frequency of the particle is πrad/s\pi rad /s,then it's amplitude is
    Solution
    Given,Diplacement  function  of  the  particle  is,x(t)=Acos (ωt+ϕ ) x(0)=1  cm  and   v(0)  =π  cmsec   differentiating  w.r.  to  t,v(t)=A(ω )sin (ωt+φ )..........(1) At  t=0,x(0)=Aωsin (ϕ )..........(2) v(0)=Aωsin (ϕ )............(3) Dividing  (3)  by  (2),v(0) x(0) =ωtan ϕ π=ωtan ϕ tan ϕ=1i.e   ϕ=3π 4  or  7π 4 sin ce  ,at   t=0x(0)=Acos (ϕ )=1cos ϕ   is   positive.Therefore  ,ϕ=7π 4 Acos (7π 4 )=1A=2  cm\begin{array}{l} Given, \\ Diplacement\, \, function\, \, of\, \, the\, \, particle\, \, is, \\ x\left( t \right) =A\cos  \left( { \omega t+\phi  } \right)  \\ x\left( 0 \right) =1\, \, cm\, \, and\, \, \, v\left( 0 \right) \, \, =\pi \, \, \dfrac { { cm } }{ { \sec   } }  \\ differentiating\, \, w.r.\, \, to\, \, t, \\ v\left( t \right) =-A\left( \omega  \right) \sin  \left( { \omega t+\varphi  } \right) ..........\left( 1 \right)  \\ At\, \, t=0, \\ x\left( 0 \right) =A\omega \sin  \left( \phi  \right) ..........\left( 2 \right)  \\ v\left( 0 \right) =-A\omega \sin  \left( \phi  \right) ............\left( 3 \right)  \\ Dividing\, \, \left( 3 \right) \, \, by\, \, \left( 2 \right) , \\ \dfrac { { v\left( 0 \right)  } }{ { x\left( 0 \right)  } } =-\omega \tan  \phi  \\ \Rightarrow \pi =-\omega \tan  \phi  \\ \Rightarrow \tan  \phi =-1 \\ i.e\, \, \, \phi =\dfrac { { 3\pi  } }{ 4 } \, \, or\, \, \dfrac { { 7\pi  } }{ 4 }  \\ \sin  ce\, \, ,at\, \, \, t=0 \\ x\left( 0 \right) =A\cos  \left( \phi  \right) =1 \\ \Rightarrow \cos  \phi \, \, \, is\, \, \, positive. \\ Therefore\, \, ,\phi =\dfrac { { 7\pi  } }{ 4 }  \\ \therefore A\cos  \left( { \dfrac { { 7\pi  } }{ 4 }  } \right) =1 \\ \Rightarrow A=\sqrt { 2 } \, \, cm \end{array}
  • Question 4
    1 / -0
    Consider the following two equations 
    A)L=IωL=I\omega and 
    B) dLdt=Γ \dfrac { dL }{ dt } =\Gamma . In noninertial frames :

    Solution
    Angular momentum L=IwL = Iw  
    A rigid object's angular momentum is stated as follows :

    • The cross product of the moment of inertia and angular velocity.
    • It is considered as analogous to corresponding linear momentum.
    • If there are no other external torque acts on the object, it is subject to the basic limitations of the angular momentum theory.
               dLdt\dfrac{dL}{dt}= Total external force torque not equal to any torque.

    So,     dLdt\dfrac{dL}{dt} = Γ is not correct.

    Hence, option (b)  A is true but B is false.  


  • Question 5
    1 / -0
    If x=θsin(α+π6)\theta sin(\alpha + \dfrac{\pi}{6}) and x1=θcosαx^1 = {\theta}cos\alpha,then what is the phase difference between the two waves.
  • Question 6
    1 / -0
    The phase (at a time t) of a particle in simple harmonic motion tells
  • Question 7
    1 / -0
    The displacement of an object attached to a spring and executing simple harmonic motion is given by x=2×102cosπtx=2\times { 10 }^{ -2 }\cos { \pi t } meters. The time at which the maximum speed first occurs is :
  • Question 8
    1 / -0
    A particle executes a SHM of time T, find the time taken by the particle to go directly from its mean position to half the amplitude
    Solution

  • Question 9
    1 / -0
    A boat has green light of wavelength λ=500nm\lambda= 500nm on the MAST. What wavelength would be measured and what colour would be observed for this light as seen by a diver submerged in water by the side of the boat?
    Given ηw=4/3\eta_{w} = 4/3 
     
  • Question 10
    1 / -0
    Speed of wave 'v' is given by
    Solution

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