Two particle executing SHM of same frequency meet at $x = + A / 2$ , while moving in opposite direction. Phase difference between the particles is ( $A \to$ Amplitude of both)

\frac{π}{6}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{5 π}{6}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{- 2 π}{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is A $\frac{- 2 π}{3}$
For particle 1

$x_{1} = a . sin (w t)$ ...(1)

For particle 2

$x_{2} = a . sin (w t + y)$ ...(2)

For instance, $x_{1} = x_{2} = \frac{a}{2}$

From (1)

$\frac{a}{2} = a . sin (w t)$

Solving we get,

$w t = \frac{π}{6}$ or $w t = \frac{5 π}{6}$

Putting these value in (2),

$sin \frac{π}{6} = a . sin (\frac{5 π}{6} + y) = \frac{a}{2}$

Solving this,

$y = - \frac{2 π}{3}$

Suggest Corrections

Similar questions

Q. Two particles executing SHM of the same amplitude and frequency on parallel lines side by side. They cross one another when moving in opposite directions each time their displacement is

\frac{\sqrt{3}}{2}

times their amplitude. The phase difference between them is:

Q. Two particles execute SHM along a straight line with same amplitude and same frequency. Everytime they meet in opposite directions at a displacement equal to half the amplitude. The difference between them is:

Q. Two particles execute SHM of same amplitude and frequency on straight lines. They pass one another when moving in opposite directions each time when their displacements is half of the amplitude. The phase difference between them is

At a moment in a progressive wave, the phase of a particle executing S.H.M. is $\frac{π}{3}$ . Then the phase of the particle 15 cm ahead and at the time $\frac{T}{2}$ will be, if the wavelength is 60 cm

Q. Two points are moving with simple harmonic motions of same period and amplitude along the same line. They are found to cross each other in opposite directions, always at the point for which the displacement

x = \frac{A}{2}

, where A is the amplitude. Find the “phase difference” between the two motions.

Join BYJU'S Learning Program

Two particle executing SHM of same frequency meet at x=+A/2, while moving in opposite direction. Phase difference between the particles is (A→ Amplitude of both)

The correct option is A −2π3For particle 1 x1=a.sin(wt) ...(1)For particle 2x2=a.sin(wt+y) ...(2)For instance, x1=x2=a2From (1)a2=a.sin(wt)Solving we get,wt=π6 or wt=5π6Putting these value in (2),sinπ6=a.sin(5π6+y)=a2Solving this,y=−2π3

Two particle executing SHM of same frequency meet at $x = + A / 2$ , while moving in opposite direction. Phase difference between the particles is ( $A \to$ Amplitude of both)