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

Concept: The phase of an oscillating system characterizes its instantaneous state of motion at any given time \(t\). To compare or compute the phase difference (\(\Delta \phi\)) between two Simple Harmonic Motions accurately, both displacement equations must be expressed using the same trigonometric function (either both as sine functions or both as cosine functions) with identical positive signs for their amplitudes. A useful trigonometric reduction formula to convert a cosine function to a sine function is: \[ \cos \theta = \sin \left(\theta + \frac{\pi}{2}\right) \]

Step 1: Analyzing the phase of the first given SHM equation.
The first simple harmonic wave equation is given as: \[ x_1 = A \sin \omega t \] Here, the phase angle of the first wave is: \[ \phi_1 = \omega t \quad \cdots (1) \]

Step 2: Converting the second SHM equation to a matching sine format.
The second wave equation is given in the cosine form: \[ x_2 = A \cos \omega t \] Using the reduction formula \(\cos \theta = \sin \left(\theta + \frac{\pi}{2}\right)\), let's rewrite \(x_2\): \[ x_2 = A \sin \left(\omega t + \frac{\pi}{2}\right) \] Now that it matches the sine function format of \(x_1\), we can identify its phase angle: \[ \phi_2 = \omega t + \frac{\pi}{2} \quad \cdots (2) \]

Step 3: Calculating the absolute phase difference between the two motions.
The phase difference \(\Delta \phi\) between the two oscillations is found by subtracting \(\phi_1\) from \(\phi_2\): \[ \Delta \phi = \phi_2 - \phi_1 = \left(\omega t + \frac{\pi}{2}\right) - \omega t \] \[ \Delta \phi = \frac{\pi}{2} \] This configuration perfectly matches option (C).