Chapter 15, Problem 1Q

Which of the following describe ϕ for the SHM of Fig. 15-20a:

(a) −π < ϕ < −π/2,

(b) π < ϕ < 3 π/3,

(c) −3 π/2 < ϕ < −π?

Figure 15-20

To determine

To find:

The limits which describes $\varnothing$ for SHM.

Answer to Problem 1Q

Solution:

The limits which describes $\varnothing$ for the SHM are $\pi <\varnothing <\frac{3\pi }{2} and-\pi <\varnothing <-\frac{\pi }{2}$.

Explanation of Solution

1) Concept:

We can use displacementof SHM to find the expression for x at t = 0.Then observing the graph, we can interpret the possible values of the phase angle.

2) Formulae:

$X\left(t\right)=A\mathrm{c}\mathrm{o}\mathrm{s}(\omega t+ \varnothing )$

3) Given:

The graph of x vs t is given.

4) Calculations:

Limit which describes $\varnothing$ for SHM:

We know the equation of SHM

$X\left(t\right)=A\mathrm{c}\mathrm{o}\mathrm{s}(\omega t+ \varnothing )$

Where A is the maximum displacement, $\omega$ is the angular velocity, t is time and $\varnothing$ is the phase difference.

At t = 0 we get,

$X=Acos\left(\varnothing \right)$

From the graph, we can see that at t = 0 the value of X is negative, we know that the value of cos is negative in II and III quadrant.

So the limits are

$\pi <\varnothing <\frac{3\pi }{2} and-\pi <\varnothing <-\frac{\pi }{2}$

Conclusion:

We can use the equation for displacement of SHM to find the phase angle of the oscillatory motion.