Think of the mass hanging from the spring in Q2. If one pulls the mass down by 'y' amount and lets go, how far up past the equilibrium point does the mass go before coming to a stop?
To determine how far up the mass goes past the equilibrium point before coming to a stop, we can use the concept of potential energy and the equation for simple harmonic motion.
First, let's consider the potential energy of the system. When the mass is pulled down by 'y' amount, it gains potential energy. The potential energy can be given by the equation:
PE = m * g * y,
where m is the mass, g is the acceleration due to gravity, and y is the displacement from the equilibrium position.
When the mass is released, it starts oscillating due to the restoring force of the spring. The maximum displacement from the equilibrium point during oscillation is known as the amplitude, denoted by 'A'.
Now, in simple harmonic motion, the potential energy at any point can be converted into kinetic energy. At the highest point of the oscillation, when the mass comes to a stop, all the potential energy is converted into kinetic energy.
So, let's equate the potential energy at the starting point (when pulled down) to the kinetic energy at the highest point:
PE = KE
m * g * y = (1/2) * m * v^2,
where v is the velocity at the highest point.
Now, the velocity at the highest point can be related to the amplitude using the equation:
v = √(2 * g * A),
where g is the acceleration due to gravity.
Substituting this equation into the previous one, we get:
m * g * y = (1/2) * m * (2 * g * A)^2,
Simplifying the equation, we find:
y = A^2.
Therefore, the mass will go up past the equilibrium point by the square of the amplitude (A).