A mass is attached to a vertical spring, which then goes into oscillation. At the high point of the oscillation, the spring is in the original unstretched equilibrium position it had before the mass was attached; the low point is 8.6cm below this.
What is the period of oscillation?
A mass of 0.5kg is attached to one end of a helical spring and produces an extension of 2.5cm. The mass is now set into verticals oscillation of amplitude 10mm.determine
1.the period of oscillation
2. the maximum kinetic energy of the mass (g=10m/s-2 )
To find the period of oscillation, we need to know the equation that relates the period to the parameters of the system, such as the mass and the spring constant. In this case, we can use Hooke's Law, which gives us the equation for the period of oscillation:
T = 2 * pi * sqrt(m/k)
Where:
T is the period of oscillation,
pi is the mathematical constant pi,
m is the mass attached to the spring, and
k is the spring constant.
In the given information, we know that the low point of oscillation is 8.6 cm below the unstretched equilibrium position. This means that the total amplitude of oscillation is 8.6 cm on either side of the equilibrium position.
However, to calculate the period, we need the spring constant (k) and the mass (m). If these values are not provided, we cannot determine the period of oscillation.