Question

what will the frequency be if 0.220 kg is added to the original mass? try to solve this problem without finding the force constant of the spring. express your answer with the appropriate units.

Answer 1

The new frequency is 0.707 times the original frequency.

Let's denote the original mass as 'm' and the added mass as 'Δm'. The initial frequency is given by:

f0 = 1/(2π) * √(k/m)

where 'k' is the spring constant. When a mass of Δm is added, the new frequency becomes:

f' = 1/(2π) * √(k/(m + Δm))

Since we are asked to solve the problem without finding the force constant of the spring, we can divide both equations by each other to eliminate 'k':

f'/f0 = √(m/(m + Δm))

Plugging in the values of m = 0.220 kg and Δm = 0.220 kg, we get:

f'/f0 = √(0.220/(0.220 + 0.220)) = 0.707 Hz

Therefore, the new frequency is 0.707 Hz times the original frequency.