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Oscillations Test - 70

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Oscillations Test - 70
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  • Question 1
    1 / -0
    Maximum tension to the minimum tension in the string during oscillations is?(Consider amplitude of oscillation =$$ \theta $$)
    Solution
    At extrem position velocity of ball will be zero.
    $$T_{min}=Mg\cos\theta$$

    For velocity at mean position
    Form conservation of energy:
    Gain in K.E = Loss in PE
    $$\dfrac{1}{2}mv^2=mg(l-l\cos\theta)$$

    $$\dfrac{mv^2}l=2mg(1-\cos\theta)$$

    $$T=mg+\dfrac{mv^2}{l}=mg+2gm(1-\cos\theta)$$

    $$T_{max}=mg[3-2\cos\theta]$$

    $$\dfrac{T_{max}}{T_{min}}=\dfrac{3-2\cos\theta}{\cos\theta}=\dfrac{3-2\left[1-2\sin^2\dfrac{\theta}{2}\right]}{1-2\sin^2\dfrac{\theta}{2}}$$

    For small angular displacement
    $$\dfrac{T_{max}}{T_{min}}=\dfrac{3-2\left[1-\dfrac{\theta^2}{2}\right]}{1-\dfrac{2\theta^2}{4}}=\dfrac{1+\theta^2}{1-\dfrac{\theta^2}{2}}$$

  • Question 2
    1 / -0

    A particle rests in equilibrium under two forces of repulsion whose centres are at a distance of $$a$$ and $$b$$ from the particle. The forces vary as the cube of the distance. The forces per unit mass are $$k$$ and $$k'$$ respectively. If the particle is slightly displaced towards one of them then, the motion is simple harmonic with the time period equal to:

    Solution

  • Question 3
    1 / -0
    The period of a conical pendulum of string length $$\sqrt 2$$ is $$2 \ s$$. The radius of the bob's orbit is about horizontal is $$[g=9.8 \ m/s^2]$$
    Solution

  • Question 4
    1 / -0
    A particle executing SHM while moving from one extreme is found at distance $$x_1,\, x_2$$ and $$x_3$$ from the centre at the end of three successive seconds. The time period of oscillation is: Here  $$\theta  = {\cos ^{ - 1}}\left( {\frac{{{x_1} + {x_3}}}{{2{x_2}}}} \right)$$
    Solution

  • Question 5
    1 / -0
    Three measurements of the time for $$20$$ oscillations of a pendulum give $$t_1 = 39.6_s , t_2 = 39.9_s and t_3 = 39.5_s$$. The precision in the measurement, is  
    Solution

  • Question 6
    1 / -0
    A particle of mass 0.1 kg executes SHM under a force F= (-10x)N. Speed of particle at mean position is 6 m/s. Then amplitude of oscillations is
    Solution

  • Question 7
    1 / -0
    If two wires of same length $$\ell $$ and area of cross section A with young modulus y and 2y connect in series and one end is fixed on roof and other end with mass m. Make simple harmonic motion, then the time period is-
    Solution

  • Question 8
    1 / -0

    A linear harmonic oscillator of force constant $$6 \times 10^5 N/m$$ and amplitude $$4 $$ cm, has total energy $$600 J$$. Select the correct statement.


    Solution

  • Question 9
    1 / -0
    Equation of SHM is x = 10 sin 10 $$\pi t$$. Find the distance between the two points where speed is $$50\pi$$ m/s. x is in cm and t is in seconds. 
    Solution

  • Question 10
    1 / -0
    A  particle executes SHM  along the line AB . IFC divided AB in the ratio 3 : 1, the ration of the time taken to travel AC and CB is : 
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