A fisherman's scale stretches 2.1 cm when a 2.3-kg fish hangs from it. What will be the frequency of vibration if the fish is pulled down 3.8 cm more and released so that it vibrates up and down?

if you answer please show your work.

F = k x

F = m g
x = 0.021 meter
m g = 2.3 * 9.81
so
k (0.021) = 2.3 * 9.81

then
omega = 2 pi f = sqrt (k/m)

the 3.8 cm has nothing to do with it.

force=kx

2.3*9.8N=k*.021m
solve for k
period=2PI sqrt(mass/k) and period= 1/freq so
freq= sqrt(k/mass) / 2PI

To find the frequency of vibration, we need to apply the formula:

Frequency (f) = 1 / Period (T)

The period is the time it takes for one complete cycle of vibration.

First, let's find the spring constant (k) of the fisherman's scale. We can use Hooke's law:

F = kx

where F is the force applied to the spring, k is the spring constant, and x is the displacement.

In this case, the force acting on the spring is the weight of the fish, which is given by:

Weight (W) = mass (m) * acceleration due to gravity (g)

W = 2.3 kg * 9.8 m/s^2 = 22.54 N

The displacement (x) of the spring is given as 2.1 cm = 0.021 m.

Therefore, we can rearrange Hooke's law to solve for the spring constant:

k = F / x = 22.54 N / 0.021 m = 1073.81 N/m

Now, let's find the new frequency of vibration when the fish is pulled down 3.8 cm more and released.

The new displacement (x) of the spring is given as 2.1 cm + 3.8 cm = 5.9 cm = 0.059 m.

To find the new force (F'), we consider the equivalent force acting on the spring, which is the weight of the fish plus the additional force applied:

F' = W + Force applied

Since the additional force applied is in the opposite direction of the weight, it subtracts from the weight. The additional force is given by:

Additional force = k * additional displacement

The additional displacement is 3.8 cm = 0.038 m.

Therefore, the additional force = 1073.81 N/m * 0.038 m = 40.87 N.

Hence, the new force becomes:

F' = 22.54 N - 40.87 N = -18.33 N

Lastly, calculating the new frequency:

k = F' / x = -18.33 N / 0.059 m = -310.85 N/m

Using the formula for frequency:

f' = 1 / T'

We need to find the new period (T').

To find the period, we use the relation between the spring constant and the period:

T = 2π * √(m / k)

For the initial situation, we have:

T = 2π * √(2.3 kg / 1073.81 N/m)

T = 2.56 s (rounded to two decimal places)

Now, for the new situation:

T' = 2π * √(2.3 kg / -310.85 N/m)

T' = 0.43 s (rounded to two decimal places)

Finally, we can calculate the new frequency:

f' = 1 / T' = 1 / 0.43 s ≈ 2.33 Hz

So, the frequency of vibration when the fish is pulled down 3.8 cm more and released is approximately 2.33 Hz.