Two bodies performing SHM have same amplitude and frequency. Their phases at a certain instant are as shown in the figure. The phase difference between them is

Question

Two bodies performing SHM have same amplitude and frequency. Their phases at a certain instant are as shown in the figure. The phase difference between them is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-s5siaic8fk9i9z3e.jpg" /> (A) (2 π /3) (B) π  (C) (π /3) (D) None of these

Tardigrade Admin · Accepted Answer

Displacement from mean position =A-((2 - √3/2))A=(√3/2)A So phase of the particle which is going to text wards +x text is φ1=(π /3) text phase difference =(5 π /3)-(π /3) =(4 π /3) text or (2 π - (4 π /3)) =(4 π /3) text or (2 π /3)

Q. Two bodies performing SHM have same amplitude and frequency. Their phases at a certain instant are as shown in the figure. The phase difference between them is

$\frac{2 π}{3}$

$π$

$\frac{π}{3}$

None of these

Displacement from mean position
$= A - (\frac{2 - 3}{2}) A = \frac{3}{2} A$ So phase of the particle which is going to
$wards + x is ϕ_{1} = \frac{π}{3}$
$phase difference = \frac{5 π}{3} - \frac{π}{3}$
$= \frac{4 π}{3} or (2 π - \frac{4 π}{3})$
$= \frac{4 π}{3} or \frac{2 π}{3}$

Q. Two bodies performing SHM have same amplitude and frequency. Their phases at a certain instant are as shown in the figure. The phase difference between them is

32π​

π

3π​

None of these

Displacement from mean position =A−(22−3​​)A=23​​A So phase of the particle which is going to wards +x is ϕ1​=3π​ phase difference =35π​−3π​ =34π​ or (2π−34π​) =34π​ or 32π​

$\frac{2 π}{3}$

$π$

$\frac{π}{3}$

Displacement from mean position
$= A - (\frac{2 - 3}{2}) A = \frac{3}{2} A$ So phase of the particle which is going to
$wards + x is ϕ_{1} = \frac{π}{3}$
$phase difference = \frac{5 π}{3} - \frac{π}{3}$
$= \frac{4 π}{3} or (2 π - \frac{4 π}{3})$
$= \frac{4 π}{3} or \frac{2 π}{3}$