Question

A pendulum of 1 metre length is hung from high building and it is freely moving to and fro like simple harmonic oscillator. Find the acceleration of simple pendulum when it crosses mean position. (amplitude of $S.H.M=5cm$)

• A. Zero
• B. $25\times {10}^{-3}m/s^2$
• C. $50\times {10}^{-3}m/s^2$
• D. $\pi \times {10}^{-2}m/s^2$

Answer 1

The correct option is B $25\times {10}^{-3}m/s^2$

We know,
$v_{max}=A\omega$
Tangential acceleration, $a_t=0$
Centripetal acceleration, $a_c=\frac{v^2}{r}=\frac{(A\omega {)}^2}{r^2}=\frac{(5cm\times \omega {)}^2}{1}$ ...(i)

$\omega =\frac{2\pi }{2\pi \sqrt{\frac{l}{g}}}=\sqrt{\frac{g}{l}}=\sqrt{\frac{10}{1}}=\sqrt{10}\text{rad/s}$
Putting this in (i)

$a_c=25\times {10}^{-4}\times 10$
$=25\times {10}^{-3}m/s$