Chapter 15: Q20P (page 436)

Figure (a)is a partial graph of the position function $x (t)$ for a simple harmonic oscillator with an angular frequency of $1 .20 rad / s$ ; Figure (b) is a partial graph of the corresponding velocity function $v (t)$ . The vertical axis scales are set by $x_{s} = 5 .0 cm$ and $v_{s} = 5 .0 cm / s$ . What is the phase constant of the SHM if the position function $x (t)$ is in the general form $x = x_{m} cos (ω t +)$ ?

Short Answer

Expert verified

The phase constant of the SHM, if the position function $x (t)$ is in the form $x (t) = x_{m} \cos (ω t + f)$ , is $- 0.695 rad$ .

Step by step solution

Stating the given data

Angular frequency of the harmonic oscillator, $ω = 1.20 rad / s$
Vertical axis scale values, $x_{s} = 5.0 cm / s$ and. $v_{s} = 5.0 cm / s$

Understanding the concept of simple harmonic motion

Using the formula of velocity function and position function, we can find the phase constant of SHM by taking the ratio of velocity and position functions.

Formulae:

The general expression for velocity of motion, $x = x_{m} cos (ω t + f)$ (i)

The general expression for velocity of motion, $v = - x_{m} ω sin (ω t + f)$ (ii)

Calculation of phase constant

Dividing equations(ii) by (i), we get

$\frac{v (t)}{x (t)} = - \frac{x_{m} ω sin (ω t + f)}{x_{m} cos (ω t + f)}$

At
$t = 0 s, v_{0} = 5 cm / s, x_{0} = 5 cm$

$\begin{matrix} \frac{v_{0}}{x_{0}} = - ω \frac{sin (f)}{cos (f)} \\ \frac{v_{0}}{- ω x_{0}} = tan f \\ f = {tan}^{- 1} (\frac{v_{0}}{- ω x_{0}}) \\ = {tan}^{- 1} [\frac{(5 cm / s)}{(- 1.20 rad / s) (5 cm)}] \\ = - 0.695 rad \end{matrix}$