A fisherman’s scale stretches 3.05 cm when a 2.5-kg fish hangs from it. What will be the frequency of vibration if the fish is pulled down 13.60 cm more and released so that it vibrates up and down?
k=mg/x=2.5•9.8/0.0305 =803.3 N/m
the natural frequency is
ω =sqrt(k/m) = sqrt(803.3/2.5)=17.9 rad/s
To determine the frequency of vibration of the fish, we can use the formula for the period of vibration:
T = 2π√(m/k)
where T is the period, m is the mass, and k is the spring constant.
First, we need to find the spring constant (k) of the fisherman's scale. The spring constant can be calculated using Hooke's Law:
F = kx
where F is the force, k is the spring constant, and x is the displacement.
In this case, the force (F) is the weight of the fish hanging from the scale, which is given by:
F = m * g
where m is the mass of the fish and g is the acceleration due to gravity (approximately 9.8 m/s^2).
So, F = 2.5 kg * 9.8 m/s^2 = 24.5 N
Next, we can calculate the spring constant (k) by rearranging Hooke's Law:
k = F / x
where F is the force and x is the displacement.
In this case, the displacement (x) is given as 3.05 cm.
Converting the displacement to meters:
x = 3.05 cm * (1 m / 100 cm) = 0.0305 m
Therefore, the spring constant (k) is:
k = 24.5 N / 0.0305 m = 803.28 N/m
Now, we can calculate the period (T) using the formula:
T = 2π√(m/k)
In this case, the mass (m) is the mass of the fish, which is given as 2.5 kg. Substituting the values into the formula:
T = 2π√(2.5 kg / 803.28 N/m) ≈ 1.78 seconds
Finally, to find the frequency (f), we can use the relationship:
f = 1 / T
Substituting the value for T:
f = 1 / 1.78 s ≈ 0.56 Hz
Therefore, the frequency of vibration when the fish is pulled down further and released will be approximately 0.56 Hz.