.35 meters in half a period, T/2
2 A = .35
x = A sin 2 pi f t
x = A sin (2 pi t/T)
max at T/4
min at 3T/4
3T/4 - T/4 = time between max and min = T/2
so
0.30 m/s = 0.35 m /(T/2)s
T/2 = 0.35/.3
T = 7/3 seconds
w = ang freq = 2pi/T = 6 pi/7
max a = w^2 A = (6pi/7)^2(.35/2)
max v = w A you do it
A 300g particle oscillating in SHM travels 35cm between the two extreme points in its motion with an average speed of 30cm/s. Find:
a) The angular frequency.
b) The maximum force on the particle.
b) The maximum speed.
1 answer
1 answer