Answer to Question #271610 in Physics for jay

Question

A tuning fork is struck once and oscillates in simple harmonic motion at a

frequency of 642 Hz and the tip of each prong moves 2.5 mm to either side of

the centre.

i)                 Calculate the maximum velocity of one of the prong’s tip in mms^-1. [ANS = 10,085mms^-1] 

ii)              Find the maximum acceleration of one of the tip of the prongs in mms^-2.

[ANS =4.07 X 10⁷ mms^-2]

Answer 1

i) The maximum velocity is

$$v_\text{max}=\omega A=2\pi fA=10085\text{ mm/s}.$$

ii) The maximum acceleration is

$$a_\text{max}=\omega^2A=(2\pi f)^2A=40.68·10^6\text{ mm/s}^2.$$