Posted by Sarah on .
A spring of negligible mass stretches 2.14 cm from its relaxed length when a force of 8.40 N is applied. A 1.300 kg particle rests on a frictionless horizontal surface and is attached to the free end of the spring. The particle is displaced from the origin to x=5.00 cm and released from rest at t=0.
--> When working this problem you calculator should be in radians mode <--
(a) What is the force constant of the spring? (b1) What is the angular frequency, (b1) What is the frequency,
Physics URGENT!!! -
k=F/Δx=8.4/0.0214 = 393 N/m
ω= sqrt(393/1.3)=302.3 rad/s
f= ω/2π=302.3/2•3.14 =48.14 Hz