A lineaer spring having a force constant k=40 N/m hangs vertically,supporting a 0.8 kg mass.the mass is then pulled down a distance of 0.15 m . During up and down oscillations ,the maximum speed of the mass will be..
6.02 m/s
A verticle spring with a force constant of 40 N/m has a mass of 0.20 kg attached to one ned, the other end being held fixed. If the mass is released and allowed to oscillate when the spring is stretched by 0.50m, what is the speed of the mass when the spring is compressed by 0.10m?
To find the maximum speed of the mass during up and down oscillations, we need to apply the principles of simple harmonic motion.
First, let's determine the spring's potential energy when the mass is pulled down a distance of 0.15 m. The potential energy of a spring can be calculated using the formula:
Potential energy (U) = (1/2) * k * x^2
where k is the force constant and x is the displacement from the equilibrium position.
Substituting the values:
k = 40 N/m and x = 0.15 m
U = (1/2) * 40 N/m * (0.15 m)^2
U = 0.225 J
The potential energy is converted into kinetic energy when the mass is at its maximum speed. Therefore, the maximum speed can be determined by equating the potential energy to the kinetic energy.
Kinetic energy (K) = (1/2) * m * v^2
where m is the mass and v is the maximum velocity.
Substituting the values:
m = 0.8 kg and U = K = 0.225 J
0.225 J = (1/2) * 0.8 kg * v^2
Simplifying the equation:
0.225 J = 0.4 kg * v^2
v^2 = (0.225 J) / (0.4 kg)
v^2 = 0.5625 m^2/s^2
Taking the square root of both sides gives:
v = √(0.5625 m^2/s^2)
v ≈ 0.75 m/s
Therefore, the maximum speed of the mass during up and down oscillations will be approximately 0.75 m/s.