a fisherman's scale stretches 3.6 cm when a 2.7 kg fish hangs from it.
what will be the amplitude and frequency of vibration if the fish is pulled down 2.5 cm more and released so that it vibrates up and down?
i think the frequency is 2.63 Hz, but i don't know how to find the amplitude
THe amplitude will be the initial deviation from the rest position, given as 2.5cm
To find the amplitude of the vibration, you need to consider the total displacement of the fish from its equilibrium position. In this case, the fish is pulled down 2.5 cm more from its initial position.
The total displacement of the fish during vibration is the sum of the stretch caused by the weight of the fish and the additional 2.5 cm displacement. Therefore, the total displacement is 3.6 cm + 2.5 cm = 6.1 cm.
Hence, the amplitude of the vibration is 6.1 cm.
Now, let's calculate the frequency of the vibration:
The formula for the frequency of vibration is given by:
Frequency (f) = 1 / Period (T)
The period (T) is the time taken for one complete cycle of vibration. In this case, it is the time taken for the fish to go from one extreme position to the other and back to its initial position.
To find the period, we need to use Hooke's law which states that the force applied to a spring is proportional to the displacement it undergoes. Mathematically, it can be expressed as:
F = -kx
where F is the force applied, k is the spring constant, and x is the displacement.
In this case, the displacement is 6.1 cm and the force is the weight of the fish, which can be calculated using the formula:
Weight = mass × acceleration due to gravity
Weight = 2.7 kg × 9.8 m/s^2 (approximating gravity as 9.8 m/s^2)
Weight = 26.46 N
Using Hooke's law, we can write:
26.46 N = k × 6.1 cm
Now, we need to convert the displacement from centimeters to meters:
6.1 cm = 0.061 m
So, the equation becomes:
26.46 N = k × 0.061 m
Solving for k:
k = 26.46 N / 0.061 m
k ≈ 433.11 N/m
Now, we can substitute the spring constant (k) value into the formula for the period:
T = 2π√(m/k)
where m is the mass of the fish, which is 2.7 kg.
T = 2π√(2.7 kg / 433.11 N/m)
T = 2π√(0.0062 kg·m/N)
Simplifying further:
T ≈ 2π × 0.0788 s
T ≈ 0.494 s
Finally, we use the formula for frequency to find the frequency:
f = 1 / T
f ≈ 1 / 0.494 s
f ≈ 2.02 Hz
Therefore, the amplitude of vibration is 6.1 cm and the frequency is approximately 2.02 Hz.