Amplitude of a mass-spring system, which is executing simple harmonic motion decreases with time. If mass =500 g, Decay constant =20 g / s then how much time is required for the amplitude of the system to drop to half of its initial value ? ( ln 2=0.693)

Question

Amplitude of a mass-spring system, which is executing simple harmonic motion decreases with time. If mass =500 g, Decay constant =20 g / s then how much time is required for the amplitude of the system to drop to half of its initial value ? ( ln 2=0.693) (A) 34.65 s (B) 17.32 s (C) 0.034 s (D) 15.01 s

Tardigrade Admin · Accepted Answer

A = A 0 e - &gamma; t= A 0 e -( bt /2 m ) ( A 0/2)= A 0 e -( bt /2 m ) ( bt /2 m )= ln 2 t =(2 m / b ) ln 2=(2 × 500 × 0.693/20) t =34.65 second .

Q. Amplitude of a mass-spring system, which is executing simple harmonic motion decreases with time. If mass $= 500 g$ , Decay constant $= 20 g / s$ then how much time is required for the amplitude of the system to drop to half of its initial value ? $(ln 2 = 0.693)$

34.65 s

17.32 s

0.034 s

15.01 s

$A = A_{0} e^{- γ t} = A_{0} e^{- \frac{b t}{2 m}}$
$\frac{A _{0}}{2} = A_{0} e^{- \frac{b t}{2 m}}$
$\frac{b t}{2 m} = ln 2$
$t = \frac{2 m}{b} ln 2 = \frac{2 \times 500 \times 0.693}{20}$
$t = 34.65$ second .

Q. Amplitude of a mass-spring system, which is executing simple harmonic motion decreases with time. If mass =500g, Decay constant =20g/s then how much time is required for the amplitude of the system to drop to half of its initial value ? (ln2=0.693)