Question

What is the phase constant for SMH with $\mathbf{\text{a}}\mathbf{\left(}\mathbf{\text{t}}\mathbf{\right)}\mathbf{}\mathbf{}$given in Fig.$\mathbf{15}\mathbf{-}\mathbf{57}$ if the position function$\mathbf{\text{x}}\mathbf{\left(}\mathbf{\text{t}}\mathbf{\right)}$has the form$\mathbf{\text{x}}\mathbf{=}{\mathbf{\text{x}}}_{\mathbf{m}}\mathbf{}\mathbf{\text{cos}}\mathbf{(}\mathbf{\omega }\mathbf{\text{t}}\mathbf{+}\mathbf{}\mathbf{\varphi }\mathbf{)}\mathbf{}\mathbf{}$and${\mathbf{\text{a}}}_{\mathbf{s}}\mathbf{=}\mathbf{}\mathbf{4.0}\mathbf{\text{\hspace{0.17em}m}}\mathbf{/}{\mathbf{\text{s}}}^{\mathbf{2}}\mathbf{}\mathbf{}$?

Short answer

The phase constant for SHM with $\mathrm{a}\left(\mathrm{t}\right)$given in the figure is $1.82\text{\hspace{0.17em}rad}$.

Step-by-step solution

The given data

• The position function of the given SHM is,$\mathrm{x}={\mathrm{x}}_{\mathrm{m}}\mathrm{co}\text{s}(\mathrm{\omega t}+\mathrm{\varphi })$.

• The acceleration of a given simple harmonic motion is, ${\mathrm{a}}_{\mathrm{s}}=4.0{\text{m/s}}^{\text{2}}$.

Understanding the concept of simple harmonic motion 

In simple harmonic motion, displacement of the particle is given by the equation,

$\mathbf{x}\mathbf{=}{\mathbf{x}}_{\mathbf{m}}\mathbf{cos}\mathbf{(}\mathbf{\omega t}\mathbf{+}\mathbf{\varphi }\mathbf{)}$

Integrating this equation twice will give us the equation for acceleration.

Using the equation for the acceleration of SHM and inserting the given value for maximum acceleration and acceleration at t =0, we can find the phase constant for SHM with maximum acceleration given in the figure.

Formulae:

The expression for the acceleration equation of the body in motion,

$\mathrm{a}\left(\mathrm{t}\right)=-{\mathrm{\omega }}^2{\mathrm{x}}_{\mathrm{m}}\text{cos}(\mathrm{\omega t}+\mathrm{\varphi })$ (i)

Calculation of phase constant

The maximum acceleration of SHM is given as:

${\mathrm{\omega }}^2{\mathrm{x}}_{\mathrm{m}}=4\text{m}/{\text{s}}^2$

Hence, substituting the value in equation (i) of the acceleration of SHM, we get

$\begin{array}{l}\mathrm{a}\left(\mathrm{t}\right)=-{\mathrm{\omega }}^2{\mathrm{x}}_{\mathrm{m}}\text{cos}(\mathrm{\omega t}+\mathrm{\varphi })\\ =-4\text{cos}(\mathrm{\omega }\text{t}+\mathrm{\varphi })\left(\text{ii}\right)\end{array}$

From the figure, we can interpret that the acceleration at $\text{t}=0$is$1\text{\hspace{0.17em}m}/{\text{s}}^2.$

Then above equation (a) becomes:

$\begin{array}{c}1=-4\text{cos}\left(\mathrm{\varphi }\right)\\ \cos \left(\mathrm{\varphi }\right)=-\frac{1}{4}\\ \mathrm{\varphi }={\cos }^{-1}\left(-\frac{1}{4}\right)\\ =1.82\text{\hspace{0.17em}rad}\end{array}$

Therefore, the phase constant for SHM with $\text{a}\left(\text{t}\right)$given in the figure is $1.82\text{\hspace{0.17em}rad}$.