Oscillations Question 12
Question 12 - 01 February - Shift 2
Choose the correct length $(L)$ versus square of time period $(T^{2})$ graph for a simple pendulum executing simple harmonic motion.
(1)
(2)
(3)
(4)
Show Answer
Answer: (3)
Solution:
Formula: Time period of simple pendulum
$ T=2 \pi \sqrt{\frac{\ell}{g}} $
$T^{2}=\frac{4 \pi^{2}}{g} \times \ell$
$T^{2} \alpha \ell$
