Question

A spring stretches by 0.0247 m when a 4.78-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f

Answer 1

This question is incomplete.So complete question is given as

I have assumed value of f=6 Hz

A spring stretches by 0.0247 m when a 4.78-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f=6 Hz (assumed value)

Answer:

mass=4.194 kg

Step-by-step explanation:

x (Spring stretches)=0.0247 m

m(mass)=4.78 kg

f=6 Hz (assumed)

mass required attached this spring=?

Solution

`k=(F)/(x)\\ k=(mg)/(x)\\ k=((4.78)*(9.8))/(0.0247)\\ k=1.89*10^(3) N/m\\ m=(k)/(4\pi^(2)f^(2)   )\\ m=(1.89*10^(3))/(4\pi^(2)(6)^(2) )\\ m=4.194 kg`