91Ó°ÊÓ

• Length of simple pendulum,$L=20\mathrm{cm}\mathrm{or}0.20\mathrm{m}$.
• Suspended mass,$m=5.0\mathrm{g}\mathrm{or}0.005\mathrm{kg}$.
• Constant speed of the car on circular path,$v=70\mathrm{m}/\mathrm{s}$.
• Radius of circular path, R=50 m.

First, we have to find the effective gravitational acceleration. Using the Pythagoras theorem, we can find the value of the effective gravitational acceleration and by using this value, we can find the frequency of oscillation of a simple pendulum.

Formula:

The effective value of gravitational acceleration,$g_{eff}=\sqrt{\left(g^2+{\left(\frac{v^2}{R}\right)}^2\right)}$ (i)

The frequency of the oscillations,$f=\frac{1}{2\mathrm{Ï„}\mathrm{Ï„}}\sqrt{\frac{g_{eff}}{L}}$ (ii)

The motion of the car is circular. So, centripetal acceleration $\mathrm{Î\pm }$is in the horizontal direction and acceleration due to gravity is in the vertically downward direction as shown in the figure below:

According to the Pythagoras theorem, we get the effective gravitational acceleration $g_{eff}$as

${g_{eff}}^2=g^2+{\mathrm{Î\pm }}^2$

Hence, using equation (i) from the above Pythagoras theorem, the effective gravitational acceleration is given as:

$$\begin{gathered} g_{eff}=\sqrt{\left({\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}^2+{\left(\frac{{\left(70\mathrm{m}/\mathrm{s}\right)}^2}{50\mathrm{m}}\right)}^2\right)} \\ =\sqrt{\left({\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}^2+{\left(98m/s^2\right)}^2\right)} \\ =98.49\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Now using equation (ii) and the value of effective gravitational acceleration, frequency of oscillation is given as:

$$\begin{gathered} f=\frac{1}{\left(2×3.14\right)}×\sqrt{\frac{98.49\mathrm{m}/{\mathrm{s}}^2}{0.20\mathrm{m}}} \\ =\frac{22.19}{6.28} \\ =3.5\mathrm{Hz} \end{gathered}$$

Hence, the required value of frequency is 3.5 Hz.