Answer:
7.08 kg
Explanation:
Given:
Length of scale (x) = 11.0 cm = 0.110 m [1 cm = 0.01 m]
Range of scale is from 0 to 200 N.
Frequency of oscillation of fish (f) = 2.55 Hz
Mass of the fish (m) = ?
Now, range of scale is from 0 to 200 N. So, maximum force, the spring can hold is 200 N. For this maximum force, the extension in the spring is equal to the length of the scale. So, ![x = 0.11\ m](https://tex.z-dn.net/?f=x%20%3D%200.11%5C%20m)
Now, we know that, spring force is given as:
![F=kx\\\\k=\frac{F}{x}](https://tex.z-dn.net/?f=F%3Dkx%5C%5C%5C%5Ck%3D%5Cfrac%7BF%7D%7Bx%7D)
Where, 'k' is spring constant.
Now, plug in the given values and solve for 'k'. This gives,
![k=\frac{200\ N}{0.11\ m}=1818.18\ N/m](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B200%5C%20N%7D%7B0.11%5C%20m%7D%3D1818.18%5C%20N%2Fm)
Now, the oscillation of the fish represents simple harmonic motion as it is attached to the spring.
So, the frequency of oscillation is given as:
![f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}](https://tex.z-dn.net/?f=f%3D%5Cfrac%7B1%7D%7B2%5Cpi%7D%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%7D)
Squaring both sides and expressing it in terms of 'm', we get:
![\frac{k}{m}=4\pi^2f^2\\\\m=\dfrac{k}{4\pi^2f^2}](https://tex.z-dn.net/?f=%5Cfrac%7Bk%7D%7Bm%7D%3D4%5Cpi%5E2f%5E2%5C%5C%5C%5Cm%3D%5Cdfrac%7Bk%7D%7B4%5Cpi%5E2f%5E2%7D)
Now, plug in the given values and solve for 'm'. This gives,
![m=\frac{1818.18\ N/m}{4\pi^2\times (2.55\ Hz)^2}\\\\m=\frac{1818.18\ N/m}{256.708\ Hz^2}\\\\m=7.08\ kg](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B1818.18%5C%20N%2Fm%7D%7B4%5Cpi%5E2%5Ctimes%20%282.55%5C%20Hz%29%5E2%7D%5C%5C%5C%5Cm%3D%5Cfrac%7B1818.18%5C%20N%2Fm%7D%7B256.708%5C%20Hz%5E2%7D%5C%5C%5C%5Cm%3D7.08%5C%20kg)
Therefore, the mass of the fish is 7.08 kg.