Answer:
a.735 N/m
b.Amplitude=0.025 m
frequency=2.63 Hz
Step-by-step explanation:
We are given that
Mass of fish =2.7 kg
1 m=100 cm
x=3.6 cm=\(\frac{3.6}{100}=0.036 m\)
a.We have to find the value of k.
We know that
\(F=-kx\)
\(mg=-kx\)
\(g=9.8 m/s^2\)
\(k=\frac{mg}{x}=\frac{2.7\times 9.8}{0.036}\)
\(k=735 N/m\)
b.We have to find the amplitude and frequency of vibration if the fish pulled down 2.5 cm more and released.
Amplitude=\(\frac{2.5}{100}=0.025 m\)
Frequency=\(\frac{1}{2\pi}\sqrt{\frac{k}{m}}\)
Substitute the values then we get
Frequency=\(\frac{1}{2\pi}\sqrt{\frac{735}{2.7}}\)
We know that \(\pi=3.14\)
Frequency=\(\frac{1}{2\cdot 3.14}\sqrt{\frac{735}{2.7}}=2.63 Hz\)
Hence, frequency of vibration=2.63 Hz
