Answer to Question #161225 in Physics for Raj

Question

A spring-mass system consisting of m= 2 kg and spring constant k = 50N/m is executing simple harmonic motion with amplitude of 0.9 m.

(a) Calculate the total mechanical energy of the oscillator

(b) What is the maximum velocity of the mass

(c) For what value(s) of displacement will the oscillator have maximum kinetic energy?

(d) For what value(s) of displacement will the oscillator have maximum potential energy ?

(e) If the oscillator is subjected to a mild damping force b=0.07 ke's find the time constant

for the damped oscillator.

Answer 1

(a) The total mechanical energy of the oscillator can be found as follows:

$$ME=U+K=\dfrac{1}{2}kx^2+\dfrac{1}{2}mv^2,$$ $$ME=\dfrac{1}{2}kA^2cos^2(\omega t+\phi)+\dfrac{1}{2}mA^2sin^2(\omega t+\phi).$$

Applying trigonometric identity $cos^2\theta+sin^2\theta=1$ and $\omega=\sqrt{\dfrac{k}{m}}$, we get:

$$ME=\dfrac{1}{2}kA^2=\dfrac{1}{2}\cdot 50\ \dfrac{N}{m}\cdot(0.9\ m)^2=20.25\ J.$$

(b) The maximum value of velocity of the mass will be when $sin(\omega t+\phi)=1$:

$$v_{max}=-\omega A=0.9\ m\sqrt{\dfrac{50\ \dfrac{N}{m}}{2\ kg}}=4.5\ \dfrac{m}{s}.$$

(c) The oscillator will have the maximum kinetic energy at the equilibrium point where $x=0$.

(d) The oscillator will have the maximum potential energy at maximum displacement from the equilibrium point.

(e)

$$\tau=\dfrac{2m}{b}=\dfrac{2\cdot2\ kg}{0.07\ \dfrac{kg}{s}}=57.14\ s.$$