Question

A 170 g mass attached to a horizontal spring oscillates at a frequency of 2.0 Hz. At t = 0 s, the mass is at x = 5.0 cm and has vx = -28 cm/s. Determine the following: (a) the period (b) the angular frequency (c) the amplitude (d) the phase constant (e) the maximum speed (f) the maximum acceleration (g) the total energy (h) the position at t = 0.4 s

Answer 1

(a) 0.5 s

In a simple harmonic motion, the period is equal to the reciprocal of the frequency:

$T=\frac{1}{f}$

where f is the frequency

For this simple harmonic oscillator, the frequency is

f = 2.0 Hz

So the period is

$T=\frac{1}{2.0 Hz}=0.5 s$

(b) 12.56 rad/s

The angular frequency is given by

$\omega = 2 \pi f$

where

f is the frequency

In this problem,

f = 2.0 Hz

So the angular frequency is

$\omega = 2 \pi (2.0 Hz)=12.56 rad/s$

(d) 0.419 rad

The displacement of the system can be written as

$x(t) = A cos (\omega t+\phi)$ (1)

where A is the amplitude and $\phi$ is the phase constant.

The velocity is the derivative of the displacement:

$v(t) = x'(t) = -A\omega sin(\omega t+\phi)$ (2)

Here we know that

at t=0, x(5)=5.0 cm and v(t)=-28 cm/s. So we can rewrite the ratio (2)/(1) as

$\frac{v(t)}{x(t)}=\frac{-28 cm/s}{5.0 cm}=\frac{-\omega A sin(\phi)}{A cos(\phi)}=-\omega tan \phi$

And re-arranging the equation we can find the phase constant:

$\phi = tan^{-1} (\frac{v}{\omega x})=tan^{-1} (\frac{-28 cm/s}{(5.0 cm)(12.56 rad/s)})=0.419 rad$

(c) 5.47 cm

The displacement of the system can be written as

$x(t) = A cos (\omega t+\phi)$ (1)

at t=0, x=5.0 cm, so using the values we found for $\omega, \phi$ we can now solve the equation to find A, the amplitude:

$A=\frac{x}{cos(\omega t+\phi)}=\frac{5.0 cm}{cos(0.419 rad)}=5.47 cm$

(e) 68.7 cm/s

The maximum speed in a simple harmonic system is given by

$v=\omega A$

where in this case we have

$\omega=12.56 rad/s$

$A=5.47 cm$

Substituting the numbers into the formula, we find

$v=(12.56 rad/s)(5.47 cm)=68.7 cm/s$

(f) 862.9 cm/s^2

The maximum acceleration in a simple harmonic system is given by

$a=\omega^2 A$

where in this case we have

$\omega=12.56 rad/s$

$A=5.47 cm$

Substituting the numbers into the formula, we find

$a=(12.56 rad/s)^2(5.47 cm)=862.9 cm/s^2$

(g) 0.04 J

The total energy of the system is equal to the kinetic energy when the speed of the system is maximum: this occurs at x=0 (equilibrium position), where the elastic potential energy is zero, and all the energy is just kinetic energy:

$E=K=\frac{1}{2}mv_{max}^2$

where we have

m = 170 g = 0.170 kg is the mass

$v_{max}=68.7 cm/s = 0.687 m/s$ is the maximum speed

Substituting into the equation,

$E=K=\frac{1}{2}(0.170 kg)(0.687 m/s)^2=0.04 J$

(h) 3.65 cm

The position of the system is given by

$x(t) = A cos (\omega t+\phi)$

where we have

$\omega=12.56 rad/s$ is the angular frequency

$A=5.47 cm$ is the amplitude

$\phi = 0.419 rad$ is the phase constant

Substituting t=0.4 s, we find the position at this time:

$x = (5.47 cm) cos ((12.56 rad/s)(0.4 s)+0.419 rad)=3.65 cm$