Self Studies
Self Studies
Self Studies
Self Studies

Waves Test - 69

Result Self Studies

Waves Test - 69
  • 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
    A simple harmonic progressive wave is represented by the equation : $$ y = 8 \sin 2 \pi \left (0.1x - 2t  \right )$$ where x and y are in cm and t is in seconds. At any instant the phase difference between two particles separated by 2.0 cm in the x-direction is     [MP PMT 2000]
    Solution
    From the given equation $$k = 0.2 \pi$$
    $$\Rightarrow \frac{2 \pi} {\lambda} = 0.2 \pi \Rightarrow \lambda = 10 \space cm$$
    $$\Delta \phi = \frac{2 \pi} {\lambda} \Delta x = \frac{2 \pi} {10} \times 2 = \frac{2 \pi} {5} = 72^{0}$$
  • Question 2
    1 / -0
    The displacement of the interfering light waves are $$y_1 = 4 sin\omega t$$ and $$y_2=3 sin \left(\omega t + \dfrac{\pi}{2}\right)$$ . What is the amplitude of the resultant wave
    Solution

  • Question 3
    1 / -0
    The amplitude of a wave represented by displacement equation $$y = \dfrac{1}{\sqrt{a}} sin\omega t \pm \dfrac{1}{\sqrt{b}} cos \omega t$$will be

    Solution
    $$y = \dfrac{1}{\sqrt{a}} sin \omega t \pm \dfrac{1}{\sqrt{b}}sin \left(\omega t + \dfrac{\pi}{2}\right)$$
    Here phase differnce = $$\dfrac{\pi}{2}\therefore$$The resultant amplitude
    $$\sqrt { \left( \dfrac { { 1 } }{ \sqrt { a }  }  \right)^2 +\left( \dfrac { 1 }{ \sqrt { b }  }  \right) ^2 } =\sqrt { \dfrac { 1 }{ a } +\dfrac { 1 }{ b }  } =\sqrt { \dfrac { a+b }{ ab }  } $$
  • Question 4
    1 / -0
    The path difference between the two waves $$y_1 = a_1 \sin \left (\omega t - \frac{2 \pi x} {\lambda}  \right )$$ and $$y_2 = a_2 \cos \left (\omega t - \frac{2 \pi x} {\lambda} + \phi \right )$$            [MP PMT 1994]
    Solution
    $$y_1 = a_1 \sin \left (\omega t - \frac{2 \pi x} {\lambda}  \right )$$ 
     $$y_2 = a_2 \cos \left (\omega t - \frac{2 \pi x} {\lambda} + \phi \right ) = a_2 \sin \left (\omega t - \frac{2 \pi x} {\lambda} + \phi + \frac{\pi} {2}\right )$$
    So phase difference = $$\phi + \frac{\pi} {2}$$ and $$\Delta = \frac{\lambda} {2 \pi} \left (\phi + \frac{\pi} {2}  \right )$$
  • Question 5
    1 / -0
    If two waves having amplitudes $$2A$$ and $$A$$ and same frequency andvelocity, propagate in the same direction in the same phase, the resulting amplitude will be
    Solution
    In the same phase $$\phi = 0$$ so resultant amplitude $$= a_1 + a_2 = 2A + A = 3A$$
  • Question 6
    1 / -0
    Two waves of frequencies $$20 Hz$$ and $$30 Hz$$. Travels out from a common point. The phase difference between them after $$0.6$$ sec is
    Solution

  • Question 7
    1 / -0
    Two waves$$y_1 = A_1 sin(\omega t -\beta_1) y_2 = A_2 sin(\omega t - \beta_2)$$ Superimpose to form a resultant wave whose amplitude is [CPMT 1999]
    Solution
    Phase difference between the two waves is 

    $$\phi = (\omega t - \beta_2) = (\omega t - \beta_1) = (\beta_1 -\beta_2)$$
    $$\therefore$$ 
    Resultant amplitude $$A = \sqrt {A_1^2 +A_2^2 + 2A_1A_2 cos(\beta_1 - \beta_2 )}$$
  • Question 8
    1 / -0
    Two waves of same frequency and intensity superimpose with each other in opposite phases, then after superposition the
    Solution
    This is a case of destructive interference 
  • Question 9
    1 / -0
    The phase difference between two points separated by 0.8 m in a wave of frequency 120 Hz is $$90^{0}$$. Then the velocity of wave will be
    Solution
    Path difference $$\Delta = \dfrac{\lambda} {2 \pi} \times \phi = \frac{\lambda} {2 \pi} \times \dfrac{\pi} {2} = \dfrac{\lambda} {4}$$
    $$\therefore \Delta = 0.8 \space m \Rightarrow \dfrac{\lambda} {4 } \Rightarrow \lambda = 3.2 \space m$$
    $$\therefore n\lambda = 120 \times 3.2 = 384 \space m/s$$
  • Question 10
    1 / -0
    Two waves are represented by $$y_1 = a sin \left(\omega t +\dfrac{\pi}{6}\right)$$and $$y_2 = a cos \omega t$$ What will be their resultant amplitude

    Solution
    $$A = \sqrt{(a_1^2 + a_2^2 + 2a_1a_2 cos\phi)}$$
    Putting $$a_1 = a_2 = a$$ and $$\phi = \dfrac{\pi}{3}$$ we get, $$A = \sqrt {3}a$$
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