A mass m at the end of a spring oscillates with a frequency of 0.83 Hz. When an additional 680-g mass is added to m, the frequency is 0.60 Hz. What is the value of m?
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Given f1=0.83Hz, f2=0.60Hz, Δm=m2-m1=0.680kg
The frequency of mass-spring oscillation is:
f = (1/2π)√(k/m)
k = m(2πf)²
Then since k is constant for both trials, we have:
k = k
m1(2πf1)² = m2(2πf2)²
m1 = m2(f2/f1)²
m1 = (m1+Δm)(f2/f1)²
m1 = Δm/((f1/f2)²-1) = 0.74kg
The frequency of mass-spring oscillation is:
f = (1/2π)√(k/m)
k = m(2πf)²
Then since k is constant for both trials, we have:
k = k
m1(2πf1)² = m2(2πf2)²
m1 = m2(f2/f1)²
m1 = (m1+Δm)(f2/f1)²
m1 = Δm/((f1/f2)²-1) = 0.74kg