For a damped harmonic oscillator of mass 200 g, the values of spring constant and damping constant are, respectively, 90 N/m and 0.04 kg/s. The time taken for its amplitude of vibration of drop to half of its initial value is ( log, 2 = 0.693)

Question

For a damped harmonic oscillator of mass 200 g, the values of spring constant and damping constant are, respectively, 90 N/m and 0.04 kg/s. The time taken for its amplitude of vibration of drop to half of its initial value is ( log, 2 = 0.693) (A) 7.0 s (B) 14.2 s (C) 15.9 s (D) 26.6 s

Tardigrade Admin · Accepted Answer

The given, K=90 N / m b=0.04 kg / s log e=2=0.693 m=200 g =0.2 kg The (instantaneous) amplitude of damped oscillations is given by At=A e-b t / 2 m Let t1 be the time taken for the amplitude to drop to half of its initial value, that is at t=t1 At=A / 2 Thus, (A/2)=A e-b t1 / 2 m eb t1 / 2 m=2 (b t1/2 m)= log e 2 t1=(2 m/b) log e 2 =(2 × 0.2/0.04) log e 2=(0.4/0.04) × 0.693 =10 × 0.693=6.93 s or ≅ 7.0 s

Q. For a damped harmonic oscillator of mass 200g, the values of spring constant and damping constant are, respectively, 90N/m and 0.04kg/s. The time taken for its amplitude of vibration of drop to half of its initial value is (log,2=0.693)

7.0 s

14.2 s

15.9 s

26.6 s

Q. For a damped harmonic oscillator of mass $200 g$ , the values of spring constant and damping constant are, respectively, $90 N / m$ and $0.04 k g / s$ . The time taken for its amplitude of vibration of drop to half of its initial value is $(lo g, 2 = 0.693)$