A 3.0kg particle is in simple harmonic motion in one dimension and moves according to the equation x=(5.0 m)cos $\left[(\mathrm{\tau \tau }/3\mathrm{rad}/s)t-\mathrm{\tau \tau }/4\mathrm{rad}\right]$,with t in seconds. (a) At what value of x is the potential energy of the particle equal to half the total energy? (b) How long does the particle take to move to this position x from the equilibrium position?

• Mass of the particle, m=30kg.
• Amplitude of the motion,$x_m=5.0\mathrm{m}$ .
• Angular frequency,$\omega =\frac{\tau \tau }{3}\mathrm{rad}/\sec$.
• Phase constant, $\varphi =\frac{\tau \tau }{4}$.

By using the expression for SHM and comparing it with the given displacement equation of x, we can find mass, amplitude, and frequency. Then we can find x using the expression for potential energy and total energy. Using the value of x as a given expression and solving it for t we can find the time taken for the required position or displacement.

Formula:

The potential energy of the oscillation system, (U)=$\frac{1}{2}m{\omega }^2{x}^2$ (i)

The total energy of the system, (T) = $\frac{1}{2}m{\omega }^2{x_m}^2$ (ii)

The expression for the displacement equation, $x=x_m\cos (\omega t-\varphi )$ (iii)

It is given that the potential energy is half of the kinetic energy,

Hence, using equations (i) and (ii), we can get the required displacement of the particle as:

$\frac{1}{2}m{\omega }^2{x}^2=\frac{1}{2}\times \frac{1}{2}m{\omega }^2{x_m}^2$

role="math" localid="1657257563219" $$\begin{gathered} x=\frac{x_m}{\sqrt{2}} \\ =\frac{5}{\sqrt{2}} \\ =3.536\mathrm{m} \\ \approx 3.5\mathrm{m}\left(\mathrm{iv}\right) \end{gathered}$$

Hence, the required position of the particle is 3.5m.

Using the displacement equation and the displacement value from equation (1), we get the time taken by the particle to reach this position as:

$$\begin{gathered} \frac{5}{\sqrt{2}}=5.0\cos \left(\frac{\tau \tau }{3}t-\frac{\tau \tau }{4}\right) \\ \frac{1}{\sqrt{2}}=\cos \left(\frac{\tau \tau }{3}t-\frac{\tau \tau }{4}\right) \end{gathered}$$

$$\begin{gathered} \frac{\tau \tau }{3}t-\frac{\tau \tau }{4}={\cos }^{-1}\frac{1}{\sqrt{2}} \\ \frac{\tau \tau }{3}t-\frac{\tau \tau }{4}={\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right)+\frac{\tau \tau }{4} \\ \frac{\tau \tau }{3}t=\frac{\tau \tau }{4}+\frac{\tau \tau }{4} \\ \frac{\tau \tau }{3}t=\frac{\tau \tau }{2} \end{gathered}$$

$$\begin{gathered} t=\frac{\tau \tau }{2}t=\frac{3}{\tau \tau } \\ =\frac{3}{2} \\ t=1.5sec \end{gathered}$$

We have to find the time taken by particle to move to position x from the equilibrium position, i.e.,role="math" localid="1657258157162" $t_{\mathit{e}\mathit{q}}\mathit{-}t$, where $t_{eq}$is the instant, the particle passes through the equilibrium position.

So, we have to set x=0

$0=5.0\cos \left(\frac{\tau \tau }{3}{t}_{eq}\frac{\tau \tau }{4}\right)$

$$\begin{gathered} \cos \left(\frac{\tau \tau }{3}{t}_{eq}-\frac{\tau \tau }{4}\right)=0 \\ \frac{\tau \tau }{3}{t}_{eq}\frac{\tau \tau }{4}={\cos }^{-1}\left(0\right)=\frac{\tau \tau }{2} \\ \frac{\tau \tau }{3}{t}_{eq}=\frac{\tau \tau }{2}+\frac{\tau \tau }{4} \\ \frac{\tau \tau }{3}{t}_{eq}=\frac{3\tau \tau }{4} \\ {t}_{eq}=\frac{3\tau \tau }{4}\times \frac{3}{\tau \tau } \\ =\frac{9}{4} \\ =2.25s \end{gathered}$$

Therefore, the time taken by the particle to reach the position from the equilibrium position is given as:

$$\begin{gathered} t_{eq}-t=2.25s-1.5s \\ =0.75s \end{gathered}$$

Hence, the value of required time is 0.75s.