Discussion Forum

The path length of oscillation of simple pendulum of length 1 m is 16 cm. Its maximum velocity is (take , g= $\pi ^{2}$ m/s2)

Answer:

Explanation:

 Given,

 Length of the pendulum(l)=1 cm

 $\therefore$   amplitude (a) = path length/2=$\frac{16}{2}$=8 cm

 Acceleration due to gravity (g) = $\pi^{2} m/s^{2}$

 We know that, time period (T)= $2\pi\sqrt{\frac{l}{g}}=2\pi\sqrt{\frac{1}{\pi^{2}}}$

  =$T=\frac{2\pi}{\pi}$

 T=2s

  $\therefore$ Maximum velocity (V_max)= a $\omega$

   = $a\times\frac{2 \pi}{T}  $   $ \left( \because \omega=\frac{2\pi}{T}\right)$

  $=\frac{8\times2\times \pi}{2}=8\pi cm/s$