91Ó°ÊÓ

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

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

$\mathbf{x}\mathbf{=}{\mathbf{x}}_{\mathbf{m}}\mathbf{cos}\mathbf{(}\mathbf{Ó\neg t}\mathbf{+}\mathbf{Ï•}\mathbf{)}$

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,

$\mathrm{a}\left(\mathrm{t}\right)=−{\mathrm{Ó\neg }}^2{\mathrm{x}}_{\mathrm{m}}\text{cos}(\mathrm{Ó\neg t}+\mathrm{Ï•})$ (i)

The maximum acceleration of SHM is given as:

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

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

$\begin{array}{l}\mathrm{a}\left(\mathrm{t}\right)=−{\mathrm{Ó\neg }}^2{\mathrm{x}}_{\mathrm{m}}\text{cos}(\mathrm{Ó\neg t}+\mathrm{Ï•})\\ =−4\text{cos}(\mathrm{Ó\neg }\text{t}+\mathrm{Ï•})\left(\text{ii}\right)\end{array}$

From the figure, we can interpret that the acceleration at $\text{t}=0$is$1\text{″¾}/{\text{s}}^2.$

Then above equation (a) becomes:

$\begin{array}{c}1=−4\text{cos}\left(\mathrm{Ï•}\right)\\ \cos \left(\mathrm{Ï•}\right)=−\frac{1}{4}\\ \mathrm{Ï•}={\cos }^{−1}\left(−\frac{1}{4}\right)\\ =1.82\text{ r²¹»å}\end{array}$

Therefore, the phase constant for SHM with $\text{a}\left(\text{t}\right)$given in the figure is $1.82\text{ r²¹»å}$.