Home
Class 11
PHYSICS
Three sinusoidal displacements are appli...

Three sinusoidal displacements are applied in X-direction to a point mass in such a way that total displacement is zero and the mass comes to rest. What are the values of B and `phi` if the equations are
`x_(1) (t) = A sin 2 omega t`
`x_(2) (t) = A sin (2 omega t + (4pi)/(3))`
`x_(3) (t) = B sin (2 omega t + phi)`

Text Solution

AI Generated Solution

To solve the problem, we need to find the values of \( B \) and \( \phi \) such that the total displacement of the point mass due to the three sinusoidal displacements is zero. The equations of the displacements are given as follows: 1. \( x_1(t) = A \sin(2 \omega t) \) 2. \( x_2(t) = A \sin(2 \omega t + \frac{4\pi}{3}) \) 3. \( x_3(t) = B \sin(2 \omega t + \phi) \) ### Step 1: Write the total displacement equation The total displacement \( x(t) \) is the sum of the three displacements: ...
Promotional Banner

Topper's Solved these Questions

  • WAVES

    MODERN PUBLICATION|Exercise REVISION EXERCISES|107 Videos
  • WAVES

    MODERN PUBLICATION|Exercise MULTIPLE CHOICE QUESTIONS|88 Videos
  • WAVES

    MODERN PUBLICATION|Exercise NCERT EXEMPLAR PROBLEMS|31 Videos
  • UNITS AND MEASUREMENT

    MODERN PUBLICATION|Exercise CHAPTER PRACTICE TEST|15 Videos
  • WORK, ENERGY AND POWER

    MODERN PUBLICATION|Exercise Chapter Practice Test|16 Videos

Similar Questions

Explore conceptually related problems

The equations of two SH M's are X_(1) = 4 sin (omega t + pi//2) . X_(2) = 3 sin(omega t + pi)

What is the amplitude of the S.H.M. obtained by combining the motions. x_(1)=4 "sin" omega t cm, " " x_(2)=4 "sin"(omega t + (pi)/(3)) cm

If i_(1) = 3 sin omega t and i_(2) =6 cos omega t , then i_(3) is

x_(1) = 3 sin omega t x_(2) = 5 sin (omega t + 53^(@)) x_(3) = - 10 cos omega t Find amplitude of resultant SHM.

If i_(1) = i_(0) sin (omega t), i_(2) = i_(0_(2)) sin (omega t + phi) , then i_(3) =

Find the displacement equation of the simple harmonic motion obtained by combining the motion. x_(1) = 2sin omega t , x_(2) = 4sin (omega t + (pi)/(6)) and x_(3) = 6sin (omega t + (pi)/(3))

Find the resultant amplitude of the following simple harmonic equations : x_(1) = 5sin omega t x_(2) = 5 sin (omega t + 53^(@)) x_(3) = - 10 cos omega t

If i_(1)=3 sin omega t and (i_2) = 4 cos omega t, then (i_3) is

The resultant amplitude due to superposition of three simple harmonic motions x_(1) = 3sin omega t , x_(2) = 5sin (omega t + 37^(@)) and x_(3) = - 15cos omega t is

The ratio of amplitudes of following SHM is x_(1) = A sin omega t and x_(2) = A sin omega t + A cos omega t