Question

$FIGUREQ15.3$shows a position-versus-time graph for a particle in SHM. What are (a) the amplitude $A$, (b) the angular frequency $\omega$, and the phase constant ${\varphi }_0$$?$

Short answer

a. The basic harmonic motion's amplitude is $10\mathrm{cm}$.

b. The simple harmonic motion's angular frequency is$\pi \mathrm{rad}/\mathrm{s}$or $3.14\mathrm{rad}/\mathrm{s}$.

c. The constant of phase${\varphi }_0$is $60°$.

Step-by-step solution

Calculation of Amplitude (part a)

(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.

Calculation of angular frequency (part b)

(b)

Calculate the angular frequency using the formula.

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

$\omega =\frac{2\pi }{T}$

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 $\omega =\frac{2\pi }{T}$and solve for $\omega$.

$\omega =\frac{2\pi \mathrm{rad}}{2\mathrm{s}}$

$=\pi \mathrm{rad}/\mathrm{s}$

$\omega =3.14\mathrm{rad}/\mathrm{s}$

Calculation of phase constant (part c)

(c)

Determine the phase constant ${\varphi }_0$using the equation

The particle's equation of motion in SHM is,

$x\left(t\right)=A\cos \left(\omega t+{\varphi }_0\right)$

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 $\pi \mathrm{rad}/\mathrm{s}$for $\omega$in the equation

$x\left(t\right)=A\cos \left(\omega t+{\varphi }_0\right)$and solve for ${\varphi }_0$.

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

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

Rearrange the variables in the equation for ${\varphi }_0$.

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

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

${\phi }_0=60°$