Two linear SHM of equal amplitudes A and frequencies $ω$ and $2 ω$ are impressed on a particle along x and y-axes respectively. If the initial phase difference between them is $π / 2$ . Find the resultant path followed by the particle.

Open in App

Solution

$x = A sin ω t$ (i)
$y = A sin (2 ω t + π / 2)$
$= A cos 2 ω t$
$= A (1 - 2 {sin}^{2} ω t)$
From Eq. (i)
$sin ω t = \frac{x}{A}$
$∴$ $y = A (1 - \frac{2 x^{2}}{A^{2}})$

Suggest Corrections

Similar questions

Two linear simple harmonic motions of equal amplitude and frequency are

impressed on a particle along x and y-axis respectively. The initial phase difference between them is

\frac{π}{2}

. The resultant path followed by the particle is :

Q. Two particles execute SHM with amplitude A and 2A and angular frequency

ω

and

2 ω

respectively. At t=0, they start with some initial phase difference. At

t = \frac{2 π}{3 ω}

, they are in same phase. Their initial phase difference is

Two particles P and Q describe SHM of same amplitude a and frequency v along the same straight line. The maximum distance between two particles is $\sqrt{2}$ a. The initial phase difference between particles is:

Q. Two particles P and Q describe SHM of same amplitude a and frequency v along the same straight line. The maximum distance between the two particles is

a \sqrt{2}

. The initial phase difference between the particle is

Q. Two particles

P

and

Q

describe SHM of same amplitude

a

and frequency

f

along the same straight line. The maximum distance between two particles is

a \sqrt{2}

. The initial phase difference between the particles is:

Join BYJU'S Learning Program

Two linear SHM of equal amplitudes A and frequencies ω and 2ω are impressed on a particle along x and y-axes respectively. If the initial phase difference between them is π/2. Find the resultant path followed by the particle.

x=Asinωt (i)y=Asin(2ωt+π/2) =Acos2ωt =A(1−2sin2ωt)From Eq. (i)sinωt=xA∴ y=A(1−2x2A2)

Two linear SHM of equal amplitudes A and frequencies $ω$ and $2 ω$ are impressed on a particle along x and y-axes respectively. If the initial phase difference between them is $π / 2$ . Find the resultant path followed by the particle.

$x = A sin ω t$ (i)
$y = A sin (2 ω t + π / 2)$
$= A cos 2 ω t$
$= A (1 - 2 {sin}^{2} ω t)$
From Eq. (i)
$sin ω t = \frac{x}{A}$
$∴$ $y = A (1 - \frac{2 x^{2}}{A^{2}})$