JEE PYQ: Oscillations Question 15
Question 15 - 2021 (25 Feb 2021 Shift 2)
Two identical springs of spring constant ‘2K’ are attached to a block of mass $m$ and to fixed support. When the mass is displaced from equilibrium position on either side, it executes simple harmonic motion. Then time period of oscillations of this system is:
(1) $\pi\sqrt{\frac{m}{k}}$
(2) $\pi\sqrt{\frac{m}{2k}}$
(3) $2\pi\sqrt{\frac{m}{k}}$
(4) $2\pi\sqrt{\frac{m}{2k}}$
Show Answer
Answer: (1)
Solution
Parallel combination: $K_{eq} = 2k + 2k = 4k$. $T = 2\pi\sqrt{\frac{m}{4k}} = \pi\sqrt{\frac{m}{k}}$.
BETA
