A fisherman's scale stretches 2.8 cm. when a 2.1 kg. fish hangs from it. What will be the frequency of vibration if the fish is pulled down and released so that it vibrates up and down?
k = m•g/x = 2.1•9.8/0.028 = 735 N/m.
ω = k/m = 735/2.1 = 0.476 rad/s,
f = ω/2•π =0.476/2•π = 0.076 Hz
To find the frequency of vibration, we need to use the formula:
f = 1 / T
where f is the frequency and T is the period of vibration.
First, let's find the period of vibration. The period is the time taken for one complete vibration cycle.
To do this, we need to find the spring constant, which is a measure of how stiff the spring of the scale is. The spring constant is denoted by k.
The formula to calculate the spring constant is:
k = F / x
where F is the force applied to the spring and x is the extension or stretch of the spring.
In this case, the force applied is the weight of the fish, which is given as 2.1 kg. To convert the mass to force, we use the formula:
F = m * g
where m is the mass and g is the acceleration due to gravity (which is approximately 9.8 m/s^2).
Calculating the force:
F = 2.1 kg * 9.8 m/s^2 = 20.58 N
Next, we will calculate the spring constant:
k = 20.58 N / 0.028 m = 735 N/m
Now, we can find the period of vibration using the formula:
T = 2π √(m / k)
where π is a mathematical constant (approximately 3.14) and m is the mass of the fish.
Calculating the period:
T = 2π √(2.1 kg / 735 N/m) ≈ 2.39 s
Finally, we can calculate the frequency using the formula f = 1 / T:
f = 1 / 2.39 s ≈ 0.42 Hz
Therefore, the frequency of vibration of the fish is approximately 0.42 Hz.