Self Studies

Oscillations Test - 26

Result Self Studies

Oscillations Test - 26
  • 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
    The figure above shows the same ideal spring in three different situations.
    First without a mass hanging the spring is unstretched.
    Next with the blue mass hanging the spring stretches.
    Finally with the red mass hanging the spring stretches twice as much as it did with the blue mass.
    How much does the red mass weigh compared to the blue mass? 

    Solution
       In equilibrium position the restoring force set up in the spring is equal to the weight of block i.e.   $$ky=mg$$ ,
    where $$y=$$ stretch in spring , $$k=$$spring constant ,
    now as both the springs are not moving in any direction so they are in equilibrium position ,
      for blue mass $$m_{B}g=ky_{B}$$ ,
      for red mass    $$m_{R}g=ky_{R}$$ , we are using same springs in both positions so k will be constant ,
    but given that ,  $$2y_{B}=y_{R}$$ ,
    so                     $$m_{R}g=k\times2y_{B}=2ky_{B}$$
    therefore    $$m_{R}/m_{B}=2$$ ,
    or               $$m_{R}=2m_{B}$$

  • Question 2
    1 / -0

    Directions For Questions

    As shown above, a spring mass system is in simple harmonic motion on the frictionless surface. Spring is fixed to the wall and mass oscillate between point A and C and B is midway between A and C.

    ...view full instructions

    When the mass passes through B, it has

    Solution
    Point B represents the mean position of oscillation whereas points A and C are the extreme positions.
    Thus the mass has maximum velocity when it passes through B and also the spring is at its natural length.
    Hence the mass has only kinetic energy but no potential energy when it passes through B.
  • Question 3
    1 / -0
    A block is attached to an ideal spring undergoes simple harmonic oscillations of amplitude A. Maximum speed of block is calculated at the end of the spring. If the block is replaced by one with twice the mass but the amplitude of its oscillations remains the same, then the maximum speed of the block will
    Solution
    Maximum speed of the block         $$V_{max}    = wA$$                     where  $$w  = \sqrt{\dfrac{k}{m}}$$
    $$\implies$$    $$V_{max}   = \sqrt{\dfrac{k}{m}} A$$                        where $$k  = constant$$

    Now the mass is doubled  i.e  $$m' = 2m$$  keeping  $$A  =constant$$ 
    $$\therefore$$       $$V'_{max} = \sqrt{\dfrac{k}{2m}} A   = \dfrac{  V_{max}}{\sqrt{2}}$$
    Thus speed is decreased by a factor of  $$\sqrt{2}$$.
  • Question 4
    1 / -0
    The black graph pictured above represents the position-time graph for a spring-mass system oscillating with simple harmonic motion.
    The colored, dashed graphs represent shapes of possible velocity time graphs for the same motion.
    The vertical axis stands for position or velocity, but the scaling does not matter. The time axis is the same for all graphs.
    Which colored graph best represents the possible velocity for the mass in this spring-mass system?

    Solution
    The given displacement graph of SHM represents the displacement $$y$$ as ,
                             $$y=a \sin\omega t$$ ,
    where $$a=$$ amplitude ,
               $$\omega=$$ angular frequency ,
    therefore velocity , 
                            $$v=dy/dt=a\omega\cos\omega t$$ 
                      or   $$v=a\omega\sin(\omega t+\pi/2)$$ ,
    it is clear that velocity v is ahead of displacement by a phase angle of $$\pi/2$$ ,which is represented by green graph of velocity .
  • Question 5
    1 / -0
    Simple harmonic oscillation of a given system can be specified completely by stating its: 
    Solution
    Although waves consist of oscillation, there is no wavelength in a pure oscillation. It is there only in waves. 
  • Question 6
    1 / -0
    Simple harmonic motion (SHM) is a technical term used to describe a certain kind of idealized oscillation. A simple harmonic oscillation has
    Solution

  • Question 7
    1 / -0
    An object swinging on the end of a string forms a simple pendulum. Some students (and some texts) often cite the simple pendulum's motion as an example of SHM. That is not quite accurate because the motion is really
    Solution
    An object performing SHM moves along a straight path.
    For large amplitudes, a pendulum moves in a curved path.
  • Question 8
    1 / -0
    Simple harmonic motion (SHM) is a technical term used to describe a certain kind of idealized oscillation. Practically all the oscillations that one can see directly in the natural world are much more complicated than SHM. Why then do physicists make such a big deal out of studying SHM?
    Solution
    Simple harmonic motion provides a basis for the characterization of more complicated motions through the techniques of Fourier analysis. Here any waveform can be represented as closely as desired by the combination of a sufficiently large number of sinusoidal waves that form a harmonic series.   Fourier's theorem suggests that any periodic function can be represented as an algebraic sum of sine and cosine functions called a Fourier Series. 
  • Question 9
    1 / -0
    A spring-mass system oscillates back and forth in positive and negative directions. The graph above shows the acceleration of the mass.
    Which of the following graphs best shows the velocity of the mass?
    The time axes are scaled the same on all graphs

    Solution
    The  displacement $$y$$ in SHM is represented  as ,
                             $$y=a \sin\omega t$$ ,
    where $$a=$$ amplitude ,
               $$\omega=$$ angular frequency ,
    therefore velocity , 
                            $$v=dy/dt=a\omega\cos\omega t$$ 
                      or   $$v=a\omega\sin(\omega t+\pi/2)$$ , 
    and acceleration ,
                            $$A=dv/dt=-a\omega^{2}\sin(\omega t)=a\omega^{2}\sin(\omega t+\pi)$$
    it is clear that velocity v is behind  by a phase angle of $$\pi/2$$ ,which is represented by  graph C .
  • Question 10
    1 / -0
    Consider a thingy hanging from a spring. The system is set vibrating by pulling the thingy down below its equilibrium position and then letting it go from rest.The frequency of the oscillation is determined by

    Solution
    The frequency is given by
    $$f=\dfrac { 1 }{ T } =\dfrac { 1 }{ 2\pi  } \sqrt { \dfrac { k }{ m }  } $$
    T = time period of the oscillation 
    m= mass of the thingy 
    k = spring constant of the spring 
    So,it is clearly seen that frequency depends on properties of spring and mass thingy.
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