91Ó°ÊÓ

(a)

The maximum displacement of a particle in simple harmonic motion is its amplitude.

The maximum displacement is$10\mathrm{cm}$

$A=10\mathrm{cm}$

The amplitude of the simple harmonic motion is reduced.

(b)

Calculate the angular frequency using the formula.

The angular frequency of a simple harmonic motion (SHM) is,

$\mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{T}$

Here, localid="1648214948272" $T$is the period of the particle in SHM.

$T=2\mathrm{s}$

Substitute $2\mathrm{s}$for $T$in the equation $\mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{T}$and solve for $\mathrm{Ó\neg }$.

$\mathrm{Ó\neg }=\frac{2\mathrm{Ï€}\mathrm{rad}}{2\mathrm{s}}$

$=\mathrm{Ï€}\mathrm{rad}/\mathrm{s}$

$\mathrm{Ó\neg }=3.14\mathrm{rad}/\mathrm{s}$

(c)

Determine the phase constant ${\mathrm{Ï•}}_0$using the equation

The particle's equation of motion in SHM is,

$x\left(t\right)=A\cos \left(\mathrm{Ó\neg }t+{\mathrm{Ï•}}_0\right)$

Time $t=0\mathrm{s}$is as follows:

$x\left(t\right)=5\mathrm{cm}$

Substitute $0\mathrm{s}$for $t$,$5\mathrm{cm}$for $x\left(t\right)$,$10\mathrm{cm}$for $A$, and $\mathrm{Ï€}\mathrm{rad}/\mathrm{s}$for $\mathrm{Ó\neg }$in the equation

$x\left(t\right)=A\cos \left(\mathrm{Ó\neg }t+{\mathrm{Ï•}}_0\right)$and solve for ${\mathrm{Ï•}}_0$.

$5\mathrm{cm}=(10\mathrm{cm})\cos \left((\mathrm{Ï€}\mathrm{rad}/\mathrm{s})(0\mathrm{s})+{\mathrm{Ï•}}_0\right)$

$=(10\mathrm{cm})\cos \left({\mathrm{Ï•}}_0\right)$

Rearrange the variables in the equation for ${\mathrm{Ï•}}_0$.

${\mathrm{Ï•}}_0={\cos }^{-1}\left(\frac{5\mathrm{cm}}{10\mathrm{cm}}\right)$

$={\cos }^{-1}\left(\frac{1}{2}\right)$

${\mathrm{φ}}_0=60°$