Question

Two particles execute simple harmonic motion along the same straight line with same amplitude and same frequency. The two particles pass one another when moving in opposite directions each time at a distan of $\frac{1{\sqrt{2}}$ times the amplitude from their common mean position. The phase differen between the two particles is

• A. $30^\circ$
• B. $45^\circ$
• C. $60^\circ$
• D. $90^\circ$

Hint:

Passing in SHM:

• When particles pass each other with opposite velocities: $x_1 = x_2$, $v_1 = -v_2$.
• Use $\sin(\omega t + \phi)$ and $\cos(\omega t + \phi)$ conditions.
• Solve using basic trigonometric identities.

Answer 1

The Correct Option is D

Let both particles have amplitude $A$. Let $x_1 = A\sin(\omega t)$, and $x_2 = A\sin(\omega t + \phi)$. At the instant of crossing, $x = \frac{A}{\sqrt{2}}$.
Then, $\sin(\omega t) = \frac{1}{\sqrt{2}} \Rightarrow \omega t = \frac{\pi}{4}$; thus, $x_2 = A\sin(\frac{\pi}{4} + \phi) = \frac{A}{\sqrt{2}}$.
Also, they are moving in opposite directions $\Rightarrow$ velocities are opposite: $\cos(\omega t + \phi) = -\cos(\omega t) = -\frac{1}{\sqrt{2}}$.
So, $\sin(\frac{\pi}{4} + \phi) = \frac{1}{\sqrt{2}}, \cos(\frac{\pi}{4} + \phi) = -\frac{1}{\sqrt{2}} \Rightarrow \frac{\pi}{4} + \phi = \frac{3\pi}{4} \Rightarrow \phi = \frac{\pi}{2}$.
Thus, the phase difference is $90^\circ$.

Answer 2

Step 1: Understand the problem
Two particles perform simple harmonic motion (SHM) along the same line with the same amplitude and frequency.
They pass each other moving in opposite directions at a distance of \(\frac{1}{\sqrt{2}}\) times the amplitude from the mean position.

Step 2: Define variables and positions
Let amplitude = \(A\)
Distance from mean position where they cross = \(x = \frac{A}{\sqrt{2}}\)

Step 3: Position equations in SHM
Displacement of first particle: \(x_1 = A \sin \theta\)
Displacement of second particle: \(x_2 = A \sin (\theta + \phi)\), where \(\phi\) is the phase difference.

Step 4: Condition when they cross moving in opposite directions
At crossing, \(x_1 = x_2 = x = \frac{A}{\sqrt{2}}\)
Also, velocities should be in opposite directions:
Velocity is proportional to the cosine of phase:
\[ v_1 \propto \cos \theta, \quad v_2 \propto \cos (\theta + \phi) \]
For opposite directions, \(v_1 \times v_2 < 0\) implies:
\[ \cos \theta \cdot \cos (\theta + \phi) < 0 \]

Step 5: Use position equality
\[ \sin \theta = \sin (\theta + \phi) = \frac{1}{\sqrt{2}} \]
So, \(\sin \theta = \sin (\theta + \phi) = \frac{1}{\sqrt{2}} = \sin 45^\circ\)
This implies:
\[ \theta = 45^\circ, \quad \text{and} \quad \theta + \phi = 135^\circ \]
Therefore,
\[ \phi = 135^\circ - 45^\circ = 90^\circ \]

Step 6: Conclusion
The phase difference between the two particles is \(90^\circ\).