A fisherman's scale stretches 3.6 cm when a 2.7-kg fish hangs from it. (a) What is the spring stiffness constant (b) what will be the amplitude and frequency of vibration if the fish is pulled down 2.5cm more and released so that it vibrates up and down..

Hooke's law states that F = -kx, so
k = - F/x = - ma / x = - 2.6kg · (-9.8m/s²) / (0.041m) = 621 N/m ← (a)

(b) f = (1/2π) · sqrt(k/m) = (1/2π) · sqrt(621N/m / 2.6kg) = 2.46/s = 2.46Hz
Amplitude is simply 2.1cm

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lublana lublana · Answer 2 · 2019-10-05T19:30:30+02:00

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=[tex]\frac{3.6}{100}=0.036 m[/tex]

a.We have to find the value of k.

We know that

[tex]F=-kx[/tex]

[tex]mg=-kx[/tex]

[tex]g=9.8 m/s^2[/tex]

[tex]k=\frac{mg}{x}=\frac{2.7\times 9.8}{0.036}[/tex]

[tex]k=735 N/m[/tex]

b.We have to find the amplitude and frequency of vibration if the fish pulled down 2.5 cm more and released.

Amplitude=[tex]\frac{2.5}{100}=0.025 m[/tex]

Frequency=[tex]\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]

Substitute the values then we get

Frequency=[tex]\frac{1}{2\pi}\sqrt{\frac{735}{2.7}}[/tex]

We know that [tex]\pi=3.14[/tex]

Frequency=[tex]\frac{1}{2\cdot 3.14}\sqrt{\frac{735}{2.7}}=2.63 Hz[/tex]

Hence, frequency of vibration=2.63 Hz