91Ó°ÊÓ

$1.$The angular frequency $\mathrm{Ó\neg }$of a vibrating object is related to the period $\mathrm{Ï„}$by:

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

$2.$The period $\mathrm{Ï„}$pendulum of length $L$is given by:

$T=2\mathrm{Ï€}\sqrt{\frac{L}{g}}$

where $g$is the gravitational acceleration?

A simple pendulum's period is only determined by its length and the size of the gravitational constant. It is independent of the mass of the object suspended at its end or the vibration amplitude.

$3.$The angle formed by a pendulum's string with the vertical fluctuates sinusoidally with time as seen below.

$\mathrm{θ}\left(t\right)={\mathrm{θ}}_{\max }\cos (\mathrm{Ó\neg }t+\mathrm{Ï•})$

where ${\mathrm{θ}}_{max}$is the maximum angle made by the string, $\mathrm{Ó\neg }$is the angular frequency of the pendulum, and $\mathrm{Ï•}$is the phase constant.

• The mass of the object hanging at the end of the string is: $m=100g$
• The length of the string of the pendulum: $L=0.1m$
• The string is initially pulled $0.8°$on one side and released

We are asked to determine the time taken by the pendulum to reach $4.0°$the opposite side.

The angular frequency of the pendulum is found in Equation $\left(1\right)$:

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

Subtitle for from equation$\left(2\right)$:

$\mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{2\mathrm{Ï€}\sqrt{\frac{L}{g}}}$

$=\sqrt{\frac{g}{L}}$

Substitute numerical value:

$\mathrm{Ó\neg }=\sqrt{\frac{9.80\mathrm{m}/{\mathrm{s}}^2}{1.0\mathrm{m}}}$

$=3.13{\mathrm{s}}^{-1}$

The angle made by the string with the vertical at a time $t$is found in Equation$\left(3\right):$

$\mathrm{θ}\left(t\right)={\mathrm{θ}}_{\max }\cos (\mathrm{Ó\neg }t+\mathrm{Ï•})$

where$\mathrm{ϕ}=0$because the pendulum is initially at the maximum angle ${\mathrm{θ}}_{\max }$

$\mathrm{θ}\left(t\right)={\mathrm{θ}}_{\max }\cos \left(\mathrm{Ó\neg }t\right)$

Substitute $-4.0°$for $\mathrm{θ}\left(t\right),8.0^{\{}∘$for${\mathrm{θ}}_{max}$and $3.13{\mathrm{s}}^{-1}$for$\mathrm{Ó\neg }$to find the time $t$at which the string makes an angle $4,0°$on the opposite side:

$-4.0°=\left(8.0°\right)\cos \left[\left(3.13{\mathrm{s}}^{-1}\right)t\right]$

Rearrange solve for $t$

$-0.5=\cos \left[\left(3.13{\mathrm{s}}^{-1}\right)t\right]$

$t=\frac{{\cos }^{-1}(-0.5)}{3.13{\mathrm{s}}^{-1}}=0.669\mathrm{s}$