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 60^{}^{{}^{\circ}}.
Options
- A
- B
- C
- D
Correct answer
A.
Step-by-step solution
The resultant amplitude of two simple harmonic motions in the same direction is given by the formula:
Substituting the given values , , and \phi = 60^{}^{{}^{\circ}}:
A = \sqrt{3.0^2 + 4.0^2 + 2(3.0)(4.0)\cos 60^{}^{{}^{\circ}}}
Rounding to one decimal place, we get .
Substituting the given values , , and \phi = 60^{}^{{}^{\circ}}:
A = \sqrt{3.0^2 + 4.0^2 + 2(3.0)(4.0)\cos 60^{}^{{}^{\circ}}}
Rounding to one decimal place, we get .