JEE PYQ: Oscillations Question 5
Question 5 - 2021 (17 Mar 2021 Shift 2)
A block of mass 1 kg attached to a spring is made to oscillate with an initial amplitude of 12 cm. After 2 minutes the amplitude decreases to 6 cm. Determine the value of the damping constant for this motion. (take $\ln 2 = 0.693$)
(1) $0.69 \times 10^2$ kg s$^{-1}$
(2) $3.3 \times 10^2$ kg s$^{-1}$
(3) $1.16 \times 10^{-2}$ kg s$^{-1}$
(4) $5.7 \times 10^{-3}$ kg s$^{-1}$
Show Answer
Answer: (3)
Solution
$A = A_0 e^{-\gamma t}$. $\ln 2 = \frac{b}{2m} \times 120$. $b = \frac{0.693 \times 2 \times 1}{120} = 1.16 \times 10^{-2}$ kg/s.
BETA
