Question

If the displacement of a particle executing SHM is given by in metre, then the frequency and maximum velocity of the particle are
A. 35 Hz, 66 m/s
B. 45 Hz, 66 m/s
C. 58 Hz, 132 m/s
D. 35 Hz, 132 m/s

Answer 1

Hint: The simple harmonic motion of an object has several quantities associated with it that relate to the equation that describes its motion: x=x$_0$cos(ωt+ϕ). And we also know that frequency =$\dfrac{\omega }{{2\pi }}$ and ${v_{\max }}$=A ω use these formulae and concepts to find the correct option.

Complete Step-by-Step solution:
The given value of y = 0.30sin (220t + 0.64)
Since the standard equation of displacement of a particle executing harmonic motion is $y = A\sin (\omega y + \phi )$here A represent amplitude \[\omega \]represent angular velocity and \[\phi \] is the phase difference
Comparing the standard equation of displacement by the given equation of displacement
Therefore we got A = 0.30, \[\omega \]= 220 radian/sec and \[\phi \]= 0.64
Since \[\omega = 2\pi f\]
$ \Rightarrow $$2\pi f = 220$
$ \Rightarrow $$f = \dfrac{{220}}{{2\pi }}$
$ \Rightarrow $$f = 35Hz$
Therefore the frequency of a particle executing SHM is 35Hz.
For maximum velocity using the formula i.e. ${V_{\max }} = A\omega $
Putting the value of A and \[\omega \]
$ \Rightarrow $${V_{\max }} = 0.30 \times 220$
$ \Rightarrow $${V_{\max }} = 66m/s$
Therefore maximum velocity of a particle executing SHM is${V_{\max }} = 66m/s$.
Hence, option A is the correct option.

Note: Simple harmonic motion is the motion in which the object moves to and fro along a line. The maximum speed attained in SHM is known as ${v_{\max }}$and can be derived from v = vmax cos (ωt + φ).