The end of one of the prongs of a tuning fork that executes simple harmonic motion of frequency 875 hz has an amplitude of 0.30 mm.
(a) find the maximum acceleration of the end of the prong.
(b) find the maximum speed of the end of the prong.
(c) find the acceleration of the end of the prong when the end has a displacement of 0.1 mm.
(d) find the speed of the end of the prong at this time.

(a)9.07×10^3 rad/s^2

(b) 1.65 m/s

(c) 3.02 ×10^3 rad/s^2

(d) 0.55m/s

Explanation:

Amplitude A = 0.30mm

f = 875Hz

Angular frequency w = 2πf

= 2×3.142×875= 5498.5 rad/s

Maximum acceleration = w^2A

= 5498.5^2× 0.3×10^-3

= 9.07×10^3rad/s^2

Maximum velocity v =wA

=1.65m/s

Acceleration at the end of the prong where displacement is 0.1mm

a =w^2x

a= = 5498.5^2× 0.1×10^-3

=3.02×10^3 rad/s^2

v =wx

v = 5498.5 × 0.1×10^-3

=0.55m/s

50 inches should be the resultant vector