Concepts Of Physics MCQ Edition [Volume 1]
- Exam
- Concepts Of Physics MCQ Edition [Volume 1]
- Subject
- Physics
- Chapter
- Simple Harmonic Motion
Concepts Of Physics MCQ Edition [Volume 1] · Physics · Simple Harmonic Motion
A particle is subjected to two simple harmonic motions of the same...
A particle is subjected to two simple harmonic motions of the same time period in the same direction. The amplitude of the first motion is while that of the second is . Determine the resultant amplitude if the phase difference between the motions is 90^{}^{{}^{\circ}}.
Options
- A
- B
- C
- D
Correct answer
C.
Step-by-step solution
The resultant amplitude of two simple harmonic motions of amplitudes and with a phase difference is given by:
Given , , and \phi = 90^{}^{{}^{\circ}}.
Substituting the values:
A = \sqrt{(3.0)^2 + (4.0)^2 + 2(3.0)(4.0) \cos 90^{}^{{}^{\circ}}}
Since \cos 90^{}^{{}^{\circ}} = 0, we get:
Given , , and \phi = 90^{}^{{}^{\circ}}.
Substituting the values:
A = \sqrt{(3.0)^2 + (4.0)^2 + 2(3.0)(4.0) \cos 90^{}^{{}^{\circ}}}
Since \cos 90^{}^{{}^{\circ}} = 0, we get: