Question

One end of a string is attached to an object of mass M, and the other end of the string is secured so that the object is at rest as it hangs from the string. When the object is raised to a position X that is a height H above its lowest point and released from rest, the object undergoes simple harmonic motion. When the object passes through the equilibrium position Y, it has a speed v0.

What methods could a student use to determine the total mechanical energy E at position Y, and why?

Answer 1

v₀ =
`√(2gH)`

to determine this speed experimentally, the student must measure the height of the body as a function of time and with equation (1) he can find the speed for each point of interest

Step-by-step explanation:

In this internal exercise the student must use the conservation of mechanical energy,

Starting point. Highest point of the trajectory

Em₀ = U = m g H

Point of interest. Point at height Y

`Em_(f)` = K + U = ½ m v² + m g Y

energy is conserved

Em₀ = Em_{f}

m g H = ½ m v² + m g Y

v² = 2 g (H -Y) (1)

in this case they indicate that Y is the equilibrium position whereby Y = 0 and the velocity is v = v₀

v₀ =
`√(2gH)`

Therefore, to determine this speed experimentally, the student must measure the height of the body as a function of time and with equation (1) he can find the speed for each point of interest