Question

In a school the physics teacher decides to perform an experiment in the lab. He left an ideal spring block system to do SHM an a frictionless table.

Five students are sent in the lab one by one and asked to write the equation of the SHM.

$Ram-{\Delta }_{0}sin(\omega t+\frac{\pi }{6})$

$Himanshu-{\Delta }_{0}sin(\omega t+\frac{2\pi }{3})$

$Satra-{\Delta }_{0}cos(\omega t+\frac{2\pi }{3})$

$Avinash-{\Delta }_{0}sin(\omega t+\frac{7\pi }{6})$

$Abhishek-{\Delta }_{0}sin(\omega t+\frac{5\pi }{6})$

Anand was asked to point out whose equations were in same phase and whose were in opposite phase?

(i)Ram and Avinash(x) same phase

(ii)Abhishek &Avinash(y) Opposite phase

(iii)Ram &Satra

Satra&Avinash

• A. (ii) -y, (iv - x)
• B. (iii) -y, iv -x
• C. (iv) -y, (ii - x)
• D. (ii) -y; I - x

Answer 1

Answer: (A), (B)

(iii) -y, iv -x

Method 1 Drawing the phase on the circle.

Ram:$A_{0}sin(\omega t+\frac{\pi }{6})$

Initial phase for Ram is $\frac{\pi }{6}={30}^{\circ }$

Himanshu:- $A_{0}sin(\omega t+\frac{2\pi }{3})$

Initial phase = $2\Pi$

Sathra-$A_{0}cos(\omega t+\frac{2\pi }{3})$

$=A_{0}sin(\frac{\pi }{2}+\omega t+\frac{2\pi }{3})$ $(\because sin(g_0+\theta )=cos\theta )$

Initial phase$=\frac{\pi }{2}+\frac{2\pi }{3}=6$

Avinash = $A_{0}sin(\omega t+\frac{7\pi }{6})$

Initial phase= $\frac{7\pi }{6}$same like satra same like satra

Abhishek =

Initial phase = $=\frac{5\pi }{6}={150}^{\circ }$

This is the final diagram. Just by looking at if we can see Satra and Avinash are in the same phase and the both are in opposite phase with respect to Ram.

So option (c) and (b)

Alternate solutions:-

(1) Ram and Avinash

$A_{0}sin(\omega t+\frac{\pi }{6})$; $A_{0}sin(\omega t+\frac{7\pi }{6})$

$A\psi =\omega t+\frac{7\pi }{6}-(\omega t+\frac{\pi }{6})=\pi$ so opposite phase

(ii) Abhishek and Avinash

$A_{0}sin(\omega t+\frac{5\pi }{6})$; $A_{0}sin(\omega t+\frac{7\pi }{6})$

$A\psi =\frac{2\pi }{6}=\frac{\pi }{3}$ soneither same nor opposite

(iii) Ram and Satra

$A_0=sin(\omega t+\frac{\pi }{6})$; $A_{0}cos(\omega t+\frac{2\pi }{3})=\left(A_{0}sin\right(\frac{\pi }{2}+\omega t+\frac{2\pi }{3})$ $\Delta \psi =\frac{\pi }{2}+\omega t+\frac{2\pi }{3})-(\omega t+\frac{\pi }{6})=\pi$ so opposite

phase

(iv) Satra and Avinash

$A_0=sin(\omega t+\frac{\pi }{2}+\frac{2\pi }{3})$; $A_{0}sin(\omega t+\frac{7\pi }{6})$

$\Delta \psi =0$ same phase.