A small fly of mass 0.26 g is caught in a spider's web. The web vibrates predominately with a frequency of 4.2 Hz. Find the effective spring stiffness constant k for the web. Hint: use natural frequency of oscillations formula.
N/m
To find the spring stiffness constant (k) for the spider's web, we can use the formula for the natural frequency of oscillations. The natural frequency (ω) is related to the stiffness constant (k) and the mass of the object (m) as follows:
ω = √(k / m)
Given information:
- Mass of the fly (m) = 0.26 g = 0.26 * 10^(-3) kg
- Frequency of web vibrations (f) = 4.2 Hz
First, let's convert the frequency from Hz to radians per second (ω).
ω = 2πf
ω = 2π * 4.2
Now, let's substitute the values into the natural frequency formula and solve for k:
ω = √(k / m)
2π * 4.2 = √(k / 0.26 * 10^(-3))
Squaring both sides of the equation, we get:
(2π * 4.2)^2 = k / 0.26 * 10^(-3)
Simplifying:
(2π * 4.2)^2 * 0.26 * 10^(-3) = k
Calculating the value:
k ≈ 67.727 N/m
Therefore, the effective spring stiffness constant (k) for the spider's web is approximately 67.727 N/m.
To find the effective spring stiffness constant (k) for the spider's web, we can use the following formula for natural frequency of oscillation:
f = 1 / (2π) * √(k / m)
where:
f is the frequency of the oscillations,
k is the spring stiffness constant, and
m is the mass of the fly.
In this case, we know that the frequency of the web's vibrations (f) is 4.2 Hz, and the mass of the fly (m) is 0.26 g.
First, we need to convert the mass to kilograms, since the SI unit of mass is kg.
0.26 g = 0.26 / 1000 kg = 0.00026 kg
Now, we can rearrange the formula to isolate the spring stiffness constant (k):
k = (2π)^2 * m * f^2
Plugging in the given values:
k = (2π)^2 * 0.00026 kg * (4.2 Hz)^2
Calculating this expression will give us the value of the spring stiffness constant (k) for the spider's web.