Home
Class 12
PHYSICS
x(1) = 5 sin omega t x(2) = 5 sin ...

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

Text Solution

AI Generated Solution

To find the amplitude of the resultant simple harmonic motion (SHM) given the equations \( x_1 = 5 \sin(\omega t) \), \( x_2 = 5 \sin(\omega t + 53^\circ) \), and \( x_3 = -10 \cos(\omega t) \), we will follow these steps: ### Step 1: Convert all equations to sine form We can express \( x_3 \) in terms of sine to make it easier to combine the equations. We know that: \[ \cos(\theta) = \sin\left(\theta + 90^\circ\right) \] Thus, we can rewrite \( x_3 \): ...
Promotional Banner

Topper's Solved these Questions

  • SIMPLE HARMONIC MOTION

    RESONANCE ENGLISH|Exercise Board Level Exercise|24 Videos
  • SIMPLE HARMONIC MOTION

    RESONANCE ENGLISH|Exercise Exercise- 1, PART - I|25 Videos
  • SIMPLE HARMONIC MOTION

    RESONANCE ENGLISH|Exercise Advanced Level Problems|13 Videos
  • SEMICONDUCTORS

    RESONANCE ENGLISH|Exercise Exercise 3|88 Videos
  • TEST PAPERS

    RESONANCE ENGLISH|Exercise PHYSICS|784 Videos

Similar Questions

Explore conceptually related problems

x_(1) = 5 sin (omegat + 30^(@)) x_(2) = 10 cos (omegat) Find amplitude of resultant SHM .

x_(1) = 3 sin omega t , x_(2) = 4 cos omega t . Find (i) amplitude of resultant SHM, (ii) equation of the resultant SHM.

x_(1) = 3 sin omega t , x_(2) = 4 cos omega t Find (i) amplitude of resultant SHM, (ii) equation of the resultant SHM.

Two simple harmonic motion are represrnted by the following equation y_(1) = 40 sin omega t and y_(2) = 10 (sin omega t + c cos omega t) . If their displacement amplitudes are equal, then the value of c (in appropriate units) is

Two simple harmonic motion are represrnted by the following equation y_(1) = 40 sin omega t and y_(2) = 10 (sin omega t + c cos omega t) . If their displacement amplitudes are equal, then the value of c (in appropriate units) is

x_(1) = 3 sin omega t implies x_(2) = 4 cos omega t . Find (i) amplitude of resultant SHm, (ii) equation of the resultant SHm.

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

A particle is subjected to SHM as given by equations x_1 = A_1 sin omegat and x_2 = A_2 sin (omega t + pi//3) .The maximum acceleration and amplitude of the resultant motion are it a_("max") and A ,respectively , then

Two SHM's are represented by y_(1) = A sin (omega t+ phi), y_(2) = (A)/(2) [sin omega t + sqrt3 cos omega t] . Find ratio of their amplitudes.