Home
Class 11
PHYSICS
The resultant amplitude due to superposi...

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

17

B

21

C

13

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
C

`x_(1)=sin omegat,x_(2)=5 sin ( omegat+37^(@))`
`x_(3)=15 sin ( omegat-pi//2)`
By the phasor diagram `,`

Get the resultant of these 3 vectors as 13.
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • DAILY PRACTICE PROBLEMS

    RESONANCE ENGLISH|Exercise DPP 77|8 Videos
  • DAILY PRACTICE PROBLEMS

    RESONANCE ENGLISH|Exercise dpp 78|3 Videos
  • DAILY PRACTICE PROBLEMS

    RESONANCE ENGLISH|Exercise dpp 75|5 Videos
  • CURRENT ELECTRICITY

    RESONANCE ENGLISH|Exercise Exercise|53 Videos
  • ELASTICITY AND VISCOCITY

    RESONANCE ENGLISH|Exercise Advanced Level Problems|9 Videos

Similar Questions

Explore conceptually related problems

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

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

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))

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

Determine resultant amplitude after super position of given four waves with help of phasor diagram. y_(1) = 15 sin omega t mm,y_(2) = 9 sin (omega t -pi//2) mm,y_(3)=7 sin (omega t +pi//2)mm & y_(4) = 13 sin omega t mm .

A partical is subjucted to two simple harmonic motions x_(1) = A_(1) sin omega t and x_(2) = A_(2) sin (omega t + pi//3) Find(a) the displacement at t = 0 , (b) the maximum speed of the partical and ( c) the maximum acceleration of the partical.

Two simple harmonic motions given by, x = a sin (omega t+delta) and y = a sin (omega t + delta + (pi)/(2)) act on a particle will be

The equation of the resultant motion of the number of simple harmonic motions is E_(c)=(1+K sin omega_(2)t) sin omega_(1)t . The number of simple harmonic components is/are.

Two particle A and B execute simple harmonic motion according to the equation y_(1) = 3 sin omega t and y_(2) = 4 sin [omega t + (pi//2)] + 3 sin omega t . Find the phase difference between them.

Find the resultant amplitude and the phase difference between the resultant wave and the first wave , in the event the following waves interfere at a point , y_(1) = ( 3 cm) sin omega t , y_(2) = ( 4 cm) sin (omega t + (pi)/( 2)), y_(3) = ( 5 cm ) sin ( omega t + pi)