Answer to Question #161222 in Physics for Raj

Question

. A 1.75-kg mass attached to a horizontally mounted spring moves as function of time as

follows :

x = 4cos (1.331 t + π/5)

Where distance is measured in meters and time in seconds. Assume it to be a free oscillator.

a) What is the phase, initial phase, amplitude, frequency, angular frequency, and period of

this motion?

b) What is the equation of the velocity of this particle?

c) What is the equation of the acceleration of this particle?

d) What is the spring constant?

e) Obtain the value of the displacement, velocity and acceleration at a time t= 2s.

f) Obtain the maximum value of displacement, velocity and acceleration.

g) What is the total mechanical energy of the oscillator?

h) Obtain the kinetic energy and the potential energy at 1= 1s.​

Answer 1

(a) Phase: $1.331t+\dfrac{\pi}{5}.$

Initial phase: $\phi=\dfrac{\pi}{5}.$

Amplitude: $A=4$.

Frequency can found as follows:

$$f=\dfrac{\omega}{2\pi}=\dfrac{1.331\ \dfrac{rad}{s}}{2\pi}=0.212\ Hz.$$

Angular frequency: $\omega=1.331\ \dfrac{rad}{s}$.

Period can be found as follows:

$$T=\dfrac{1}{f}=\dfrac{1}{0.212\ Hz}=4.72\ s.$$

b)

$$v(t)=\dfrac{d}{dt}x(t),$$ $$v(t)=\dfrac{d}{dt}(4cos(1.331t+\dfrac{\pi}{5}))=-1.331\cdot4sin(1.331t+\dfrac{\pi}{5}),$$ $$v(t)=-5.324sin(1.331t+\dfrac{\pi}{5}).$$

c)

$$a(t)=\dfrac{d}{dt}v(t),$$ $$a(t)=\dfrac{d}{dt}(-5.324sin(1.331t+\dfrac{\pi}{5}))=-5.324\cdot1.331cos(1.331t+\dfrac{\pi}{5}),$$ $$a(t)=-7.1cos(1.331t+\dfrac{\pi}{5}).$$

d) The spring constant can be found as follows:

$$\omega=\sqrt{\dfrac{k}{m}},$$ $$k=\omega^2m=(1.331\ \dfrac{rad}{s})^2\cdot1.75\ kg=3.1\ \dfrac{N}{m}.$$

e)

$$x(t=2\ s)=4\ m\cdot cos(1.331\ \dfrac{rad}{s}\cdot2\ s+\dfrac{\pi}{5})=4\ m,$$ $$v(t=2\ s)=-5.324\ \dfrac{m}{s}\cdot sin(1.331 \dfrac{rad}{s}\cdot2\ s+\dfrac{\pi}{5})=-0.305\ \dfrac{m}{s},$$ $$a(t=2\ s)=-7.1\dfrac{m}{s^2}\cdot cos(1.331\ \dfrac{rad}{s}\cdot 2\ s+\dfrac{\pi}{5})=-7.08\ \dfrac{m}{s^2}.$$

f) The maximum value of displacement will be when $cos(1.331t+\dfrac{\pi}{5})=1$:

$$x_{max}=4\ m.$$

The maximum value of velocity will be when $sin(1.331t+\dfrac{\pi}{5})=1$:

$$v_{max}=-5.324\ \dfrac{m}{s}.$$

The maximum value of acceleration will be when $cos(1.331t+\dfrac{\pi}{5})=1$:

$$a_{max}=-7.1\ \dfrac{m}{s^2}.$$

g) 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:

$$ME=\dfrac{1}{2}kA^2=\dfrac{1}{2}\cdot 3.1\ \dfrac{N}{m}\cdot(4\ m)^2=24.8\ J.$$

h)

$$K(t=1\ s)=\dfrac{1}{2}\cdot3.1\ \dfrac{N}{m}\cdot(4\ m)^2cos^2(1.331\ \dfrac{rad}{s}\cdot1\ s+\dfrac{\pi}{5})=24.77\ J,$$ $$U(t=1\ s)=\dfrac{1}{2}\cdot1.75\ kg\cdot(4\ m)^2sin^2(1.331\ \dfrac{rad}{s}\cdot1\ s+\dfrac{\pi}{5})=0.016\ J.$$