JEE Main & Advanced
Physics
Simple Harmonic Motion
Question Bank
Time Period and Frequency
question_answer
The displacement x (in metres) of a particle performing simple harmonic motion is related to time t (in seconds) as \[x=0.05\cos \left( 4\,\pi \,t+\frac{\pi }{4} \right)\]. The frequency of the motion will be [MP PMT/PET 1998]
A) 0.5 Hz
B) 1.0 Hz
C) 1.5 Hz
D) 2.0 Hz
Correct Answer:
D
Solution :
From the given equitation \[\omega =2\pi n=4\pi \] Þ \[n=2Hz\]