Self Studies

Oscillations Test - 27

Result Self Studies

Oscillations Test - 27
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0
    We are still looking at the oscillating thingy hanging from a spring. The system was set vibrating by pulling the thingy down below its equilibrium position and then letting it go from rest. If the initial displacement is doubled what happens to the maximum kinetic energy of the thing?

    Solution
    Maximum kinetic energy (KE) occurs when the thingy's speed is greatest and is proportional to the square of that speed. The maximum speed occurring at the equilibrium point is given by $$A\omega$$, A being the amplitude and $$\omega$$ being the angular frequency. Since, $$\omega$$ is irrespective of displacement so, velocity depends on the amplitude and doubling A results in increasing maximum kinetic energy by a factor of 4.
  • Question 2
    1 / -0
    The displacement of a particle is represented by the equation $$y=3 cos(\dfrac{\pi}{4}- 2 \omega t)$$. The motion of the particle is :
    Solution

  • Question 3
    1 / -0
    A particle executing SHM has a maximum speed of $$0.5 m{s}^{-1}$$ and maximum acceleration of $$1 m {s}^{-2}$$. 
    The angular frequency of oscillation is:
    Solution
    Maximum velocity, $$V_{max} = Aw$$
    Maximum acceleration, $$a_{max} = Aw^2$$
    $$\implies$$ Angular frequency, $$w = \dfrac{a_{max}}{V_{max}} = \dfrac{1}{0.5} =2$$   $$rad/s$$
  • Question 4
    1 / -0
    In their discussions of SHM text books sometimes consider a thingy attached to a horizontal spring and moving horizontally on a frictionless surface, instead of the hanging thingy that we have been looking at. Suppose that the two springs and the two thingies are identical. Think about whether these two systems are significantly different in other respects and decide which one of the following statements is true.

    Solution
    The time period of the $$0\int$$ mass system is given by 
    $$T = 2\pi \sqrt { \dfrac { m }{ k }  } $$
    m = mass attached to spring, K = spring constant.
    clearly, it is seen that time period only depends on mass attached to spring and spring constant. In the given two cases both are the same no matter what the weight of the spring is.

  • Question 5
    1 / -0
    Which of  the  following regarding oscillatory motion is true?
    Solution
    Oscillator motion is to and fro motion about a mean position. Earth motion is not a to and fro motion here, hence it is not an oscillatory motion. But as earth motion is repeated in a regular interval of time, its motion is periodic.
  • Question 6
    1 / -0
    Which of the following is an example of oscillatory motion? 
    Solution
    Oscillation of a simple pendulum is a simple harmonic motion.
  • Question 7
    1 / -0
    A mass of 2.0 kg is put on a flat pan attached to a vertical spring fixed on the ground as shown in the figure. The mass of the spring and the pan is negligible. When pressed slightly and released the mass executes a simple harmonic motion. The spring constant is 200 N/m. What should be the minimum amplitude of the motion, so that the mass gets detached from the pan?
    (Take $$\displaystyle g=10{ m }/{ { s }^{ 2 } }$$)

    Solution
    Let the minimum amplitude of SHM is $$a$$.
    Restoring force on spring    $$\displaystyle F=ka$$
    Restoring force is balanced by weight $$mg$$ of block. For mass to execute simle hormonic motion of amplitude $$a$$,
    $$\displaystyle \therefore ka=mg$$
    or $$\displaystyle a=\frac { mg }{ k } $$
    Here, $$\displaystyle m=2kg,k=200{ N }/{ m },g=10{ m }/{ { s }^{ 2 } }$$
    $$\displaystyle \therefore \quad a=\frac { 2\times 20 }{ 200 } =\frac { 10 }{ 100 } m$$  $$\displaystyle =\frac { 10 }{ 100 } \times 100cm=10cm$$
    Hence, minimum amplitude of the motion should be $$10$$ cm, so that the mass gets detached from the pan.
  • Question 8
    1 / -0
    Which of the following conditions must be satisfied for a body to oscillate or vibrate?
    A: The body must have inertia to keep it moving across the mid point of its path.
    B: There must be a restoring force to accelerate the body towards the midpoint.
    C: The fractional force acting on the body against its motion must be small.
    Solution

  • Question 9
    1 / -0
    The displacement of a particle is represented by the equation $$y=sin^3(\omega t)$$. The motion is 
    Solution
    Given the equation of displacement of the particle, $$y={ sin }^{ 3 }\omega t$$
    We know $$sin3\theta =3sin\theta -4{ sin }^{ 3 }\theta $$
    Hence, $$y=\frac { (3sin\omega t-4sin3\omega t) }{ 4 } \\ \Rightarrow 4\frac { dy }{ dt } =3\omega cos\omega t-4\times [3\omega cos3\omega t]\\ \Rightarrow 4\times \frac { { d }^{ 2 }y }{ { dt }^{ 2 } } =-3{ \omega  }^{ 2 }sin\omega t+12\omega sin3\omega t\\ \Rightarrow \frac { { d }^{ 2 }y }{ { dt }^{ 2 } } =\frac { -3{ \omega  }^{ 2 }sin\omega t+12\omega sin3\omega t }{ 4 } \\ \Rightarrow \frac { { d }^{ 2 }y }{ { dt }^{ 2 } } $$ is not proportional to y. 
    Hence, the motion is not SHM. 
    As the expression is involving sine function, hence it will be periodic. 
    Also $${ sin }^{ 3 }\omega t={ \left( sin\omega t \right)  }^{ 3 }\\ ={ [sin(\omega t+2\pi )] }^{ 3 }\\ ={ [sin(\omega t+2\pi /\omega )] }^{ 3 }$$
    Hence, $$y={ sin }^{ 3 }\omega t$$ represents a periodic motion with period $$2\pi /\omega $$.
  • Question 10
    1 / -0
    The velocity  vector $$v$$ and displacement vector $$x$$ of a particle executing SHM are related as $$\dfrac{v dv}{dx} ={\omega}^2 x$$ with the initial condition $$v = v_0$$ at  $$x = 0$$. The velocity $$v$$, when displacement is $$x$$, is:
    Solution
    As it is SHM so the equation of motion will be $$F=-kx $$ or $$\dfrac{vdv}{dx}=-\omega^2 x$$
    Now integrating the expression with boundary condition, $$\int^v_{v_0}vdv=-\omega^2\int^x_0xdx$$  
    or $$\dfrac{1}{2}[v^2-v_0^2]=-\dfrac{\omega^2x^2}{2}$$
    or $$v=\sqrt{v_0^2-\omega^2x^2}$$
Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now