Question

What will be the displacement equation of the simple harmonic motion obtained by combining the motions? Given: \[ x_1 = 2 \sin(\omega t), \quad x_2 = 4 \sin(\omega t + \frac{\pi}{2}), \quad x_3 = 6 \sin(\omega t) \]

• A. \( x = 10.25 \sin(\omega t + \phi) \)
• B. \( x = 10.25 \sin(\omega t - \phi) \)
• C. \( x = 11.25 \sin(\omega t + \phi) \)
• D. \( x = 11.25 \sin(\omega t - \phi) \)

Hint:

When combining simple harmonic motions with phase differences, use vector addition and trigonometric identities to find the resultant displacement.

Answer 1

The Correct Option is C

The displacement equation of simple harmonic motion obtained by combining the motions is the result of vector addition of the displacements.
We first need to sum the two displacements: The displacements \( x_1 \) and \( x_3 \) have the same phase and can be added directly: \[ x_1 + x_3 = 2 \sin(\omega t) + 6 \sin(\omega t) = 8 \sin(\omega t) \] Next, \( x_2 \) has a phase shift of \( \frac{\pi}{2} \), and we use the trigonometric identity to combine it with the sum of the other two: \[ x_2 = 4 \sin(\omega t + \frac{\pi}{2}) = 4 \cos(\omega t) \] Now, we combine \( 8 \sin(\omega t) \) and \( 4 \cos(\omega t) \) using the trigonometric identity: \[ R \sin(\omega t + \phi) = \sqrt{8^2 + 4^2} \sin(\omega t + \phi) = 8.944 \sin(\omega t + \phi) \] The combined displacement is approximately: \[ x = 11.25 \sin(\omega t + \phi) \]
Thus, the correct answer is (c).