Posted by Hannah on .
A spring stretches by 0.0226 m when a 4.39-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 7.35 Hz?
7.35 = 0.0226(4.39x)
Did I set this up correctly?
Physics(please check) -
The spring constant is
k =mg/Δx=4.39•9.8/0.0226 =1903 N/m.
Natural frequency is
f=sqrt(k/m) sqrt (1903/4.39) = 20.8 Hz