Animated Solution for Physics - Oscillations: Two simple harmonic motions are represented by the equations $y_1=0.1\sin \left(100\pi t+\frac{\pi }{3}\right)$ and $y_2=0.1\cos \pi t$. The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is

Two Oscillating Particles

Imagine two particles dancing back and forth, each following its own unique rhythm. The problem gives us their displacement equations:

Our mission is to find the phase difference between their velocities, not their displacements. This means we first need to figure out how fast they are moving at any given instant.

Velocity is simply the rate of change of displacement. Let's differentiate the first equation with respect to time. Using the chain rule, the derivative of sine becomes cosine, and the inner coefficient $100\pi$ pops out:

Now, to compare phases accurately, we must establish a common ground. The standard convention is to express all wave equations as positive sine functions. We can convert a cosine into a sine by adding $\frac{\pi }{2}$ to the phase:

Next, we tackle the second particle. Differentiating its displacement gives:

We have a negative sign here! To absorb it and make the function a positive sine wave, we add $\pi$ to the phase:

Here is where many students stumble. If you look closely, the angular frequencies of the two particles are different ($100\pi$ versus $\pi$). Because they are oscillating at different speeds, their phase difference is technically changing every single second!

However, in standard competitive exams, when asked for "the" phase difference without a specified time, it is universally implied to calculate the initial phase difference at $t=0$.

Let's extract the initial phases at $t=0$:

For particle 1: ${\varphi }_1=\frac{5\pi }{6}$

For particle 2: ${\varphi }_2=\pi$

The phase difference of particle 1 with respect to particle 2 is simply the difference between these two values:

And there we have it! The negative sign indicates that particle 1's velocity phasor is lagging behind particle 2's velocity phasor by $30^{\circ }$ at the start of the clock.