JEE Main 2021 · 17 March · evening · Single correct · +4 / −1
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)
- (a) 0.69 × 10² kg s⁻¹
- (b) 3.3 × 10² kg s⁻¹
- (c) 1.16 × 10⁻² kg s⁻¹
- (d) 5.7 × 10⁻³ kg s⁻¹
Answer and solution
Answer (c) 1.16 × 10⁻² kg s⁻¹
Solution
Physics sir · 4 steps
Amplitude halves in 120 s, so bt/2m = ln 2. Keep the 2 in the denominator and solve for b.
- Damped amplitude. With a damping force -bv, the amplitude does not stay fixed. It decays exponentially with time, and the exponent has 2m in its denominator.
A = A₀ e^(-bt/2m)
- Use the halving. The amplitude drops from 12 cm to 6 cm, so A/A₀ = 1/2 and e^(bt/2m) = 2. Taking logs gives bt/2m = ln 2, with t = 2 min = 120 s.
- Put in numbers. b = (2m ln 2)/t = (2 × 1 × 0.693)/120 = 1.386/120 = 1.155 × 10⁻² kg s⁻¹.
- So. The damping constant is about 1.16 × 10⁻² kg s⁻¹, which is option (c).
The idea to keep. For a damped oscillator, amplitude falls as e^(-bt/2m), while energy falls as e^(-bt/m). A halving amplitude means bt/2m = ln 2. Always convert time to seconds first.
Answer from the NTA final answer key. Solution by PrepWiser.