The body mass is 0.4kg , its hung on a spring with stiffness of 90 N/m
What is the frequency of its oscillation?
Period=2PI sqrt(mass/stiffness)=2PI*sqrt(.4/90)
frequency= 1/period
To find the frequency of oscillation of the body, we can use Hooke's Law and the equation for frequency of simple harmonic motion.
Hooke's Law states that the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position. Mathematically, this can be expressed as F = -kx, where F is the force exerted by the spring, k is the spring constant (stiffness), and x is the displacement from the equilibrium position.
In this case, the body has a mass of 0.4 kg and is hung on a spring with a stiffness of 90 N/m. The force exerted by the spring can be calculated using the formula F = mg, where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s^2).
F = mg = 0.4 kg * 9.8 m/s^2 = 3.92 N
Using Hooke's Law, we can set the force exerted by the spring equal to the mass times the acceleration to find the displacement:
-kx = ma
Since a = -ω^2x (where ω is the angular frequency), we have:
-kx = m * (-ω^2x)
Rearranging the equation:
ω^2 = k/m
Substituting the values:
ω^2 = 90 N/m / 0.4 kg
Simplifying:
ω^2 = 225 rad^2/s^2
To find the angular frequency, we take the square root:
ω = √(225 rad^2/s^2) = 15 rad/s
Finally, we can use the equation for frequency in simple harmonic motion:
f = ω / (2π)
Substituting the value of ω:
f = 15 rad/s / (2π) ≈ 2.39 Hz
Therefore, the frequency of oscillation for the body is approximately 2.39 Hz.