Question

Two SHMs are given by: \[ x_1 = A \sin \omega t, \] \[ x_2 = A \cos \omega t. \] Phase difference is:

• A. \( 0 \)
• B. \( \pi / 4 \)
• C. \( \pi / 2 \)
• D. \( \pi \)

Hint:

To remember phase conversions visually, think of sine and cosine curves. A standard cosine wave starts at its maximum value at \(t=0\), while a sine wave starts at zero. Because the cosine wave reaches its peaks earlier, it always leads the corresponding sine wave by exactly a quarter cycle, which is a phase angle of \(\pi/2\) radians (\(90^\circ\))!

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Phase represents the state of motion of an oscillating particle.
To find the phase difference between two harmonic motions, they must both be expressed using the same trigonometric function (either both sine or both cosine) with positive coefficients.
Key Formula or Approach:
Use the trigonometric reduction formula:
\[ \cos \theta = \sin\left(\theta + \frac{\pi}{2}\right) \]
Step 2: Detailed Explanation:
The first equation is already in sine form:
\[ x_1 = A\sin(\omega t) \implies \text{Phase } \phi_1 = \omega t \]
The second equation is in cosine form:
\[ x_2 = A\cos(\omega t) \]
Convert \(x_2\) into a sine function to match the format of \(x_1\):
\[ x_2 = A\sin\left(\omega t + \frac{\pi}{2}\right) \implies \text{Phase } \phi_2 = \omega t + \frac{\pi}{2} \]
The phase difference \(\Delta\phi\) is the subtraction of the two phase angles:
\[ \Delta\phi = \phi_2 - \phi_1 = \left(\omega t + \frac{\pi}{2}\right) - (\omega t) = \frac{\pi}{2} \]
Alternatively, we know a cosine wave starts at its peak (\(t=0, x=A\)) while a sine wave starts at zero (\(t=0, x=0\)). The cosine wave "leads" the sine wave by exactly a quarter cycle.
Step 3: Final Answer:
The phase difference between the two given SHMs is \(\pi/2\).