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(D) The angular amplitude is $	heta_0 = \cos^{-1}(0.9)$. Since $	heta_0$ is small,$\cos 	heta_0 \approx 1 - \frac{	heta_0^2}{2} = 0.9$,which gives

Vedclass · Accepted Answer

$(31.19$ to $32.27)$

Vedclass · Answer

(D) The angular amplitude is $	heta_0 = \cos^{-1}(0.9)$. Since $	heta_0$ is small,$\cos 	heta_0 \approx 1 - \frac{	heta_0^2}{2} = 0.9$,which gives $\frac{	heta_0^2}{2} = 0.1$,so $	heta_0^2 = 0.2$ and $	heta_0 = \sqrt{0.2} = \sqrt{\frac{1}{5}}$.For a simple pendulum,the maximum velocity is $v' = \ell \omega 	heta_0 = \ell \sqrt{\frac{g}{\ell}} 	heta_0 = 	heta_0 \sqrt{g\ell}$.Substituting the values: $v' = \sqrt{\frac{1}{5}} \sqrt{10 	imes 8} = \sqrt{\frac{80}{5}} = \sqrt{16} = 4 \ m/s$.The Doppler shift formula for source and observer moving towards each other is $f_{max} = \left( \frac{v + v'}{v - v'} ight) f$ and moving away is $f_{min} = \left( \frac{v - v'}{v + v'} ight) f$.The maximum variation in frequency is $\Delta f = f_{max} - f_{min} = f \left( \frac{v + v'}{v - v'} - \frac{v - v'}{v + v'} ight) = f \left( \frac{(v + v')^2 - (v - v')^2}{v^2 - v'^2} ight) = f \left( \frac{4vv'}{v^2 - v'^2} ight)$.Substituting $v = 330 \ m/s$,$v' = 4 \ m/s$,and $f = 660 \ Hz$:$\Delta f = 660 	imes \left( \frac{4 	imes 330 	imes 4}{330^2 - 4^2} ight) = 660 	imes \left( \frac{5280}{108900 - 16} ight) \approx 660 	imes \left( \frac{5280}{108884} ight) \approx 660 	imes 0.04849 \approx 32.003 \ Hz$.Thus,the value lies in the range $(31.19$ to $32.27)$. Therefore,the correct option is $D$.

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