The scale of a spring balance reading from zero to 200 N is 12.5 cm long. A fish hanging from the bottom of the spring oscillates vertically at 2.60 Hz.
2 Answers
-
The formula for the oscillation of a mass on a spring is T=2π√(m/k)
The relationship between frequency and period is T=1/f so the period is 1/2.60=0.3846 s.
From the information given, a force of 200N will stretch the spring by 0.125m.
That enables us to find the spring constant since k=F/x
k=200/0.125=1,600 N/m
Rearrange the formula
T=2π√(m/k)
T²=4π²m/k
m=kT²/4π²
m=_____________ kg
-
__________________________________
Force constant=200/0.125=1600 N/m
Frequency= f =(1/2pi) sq rt [k/m]
squaring and simplifying,
mass = m = k /(2pi*f)^2
m=1600 /(2pi*2.6)^2=5.9953 kg
the mass of the fish is 5.9953 kg
______________________
