Question

A mass vibrates back and forth from the free end of an ideal spring of spring constant 20.0 N/m with an amplitude of 0.250 m. What is the maximum kinetic energy of this vibrating mass? What is its velocity when it is 0.125 m from its maximum displacement?

Answer 1

Answer:

The kinetic energy of this vibrating mass and the velocity are 0.625 J and $\dfrac{0.968}{\sqrt{m}}\ m/s$

Explanation:

Given that,

Spring constant = 20.0 N/m

Amplitude = 0.250 m

Displacement = 0.125 m

At equilibrium position,

The kinetic energy is maximum and potential energy is zero.

We need to calculate the kinetic energy

Using formula of maximum energy

$K.E=\dfrac{1}{2}ka^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times20.0\times(0.250)^2$

$K.E=0.625\ J$

We need to calculate the velocity

Using formula of velocity

$v=\omega\sqrt{(A^2-x^2)}$

We know that.

The spring constant $k = m\omega^2$

Put the value of $\omega$ in to the formula

$v^2=\dfrac{k}{m}\times(A^2-x^2)$

Put the value into the formula

$v^2=\dfrac{20(0.250^2-0.125^2)}{m}$

$v^2=\dfrac{0.9375}{m}$

$v=\sqrt{\dfrac{0.9375}{m}}$

$v=\dfrac{0.968}{\sqrt{m}}\ m/s$

Hence, The kinetic energy of this vibrating mass and the velocity are 0.625 J and $\dfrac{0.968}{\sqrt{m}}\ m/s$