A block of mass 200 g executing SHM under the influence of a spring of spring constant k = 90 N m-1 and a damping constant b = 40 g s-1. The time elapsed for its amplitude to drop to half of its initial value is (Given, In (l/2) = -0.693), the time elapsed for its mechanical energy to drop half of its initial value is

Question

A block of mass 200 g executing SHM under the influence of a spring of spring constant k = 90 N m-1 and a damping constant b = 40 g s-1. The time elapsed for its amplitude to drop to half of its initial value is (Given, In (l/2) = -0.693), the time elapsed for its mechanical energy to drop half of its initial value is (A) 2.5 s (B) 3.5 s (C) 4.5 s (D) 7.5 s

Tardigrade Admin · Accepted Answer

The energy of the damped oscillator at any instant t is given by

E = E_{0} e^{- b t / m} \dots (i)

where

E_{0}

is its initial energy and

b

is the damping constant
At

t = t_{1/2}

, the energy drop to half of its initial value From Eq.(i), we get

\frac{E _{0}}{2} = E_{0} e^{- b t_{1/2^{/} m}} \frac{1}{2} = e^{- b t_{1/2 /^{m}}}

Taking natural logarithm on both sides, we get
ln

(\frac{1}{2}) = - \frac{b t _{12}}{m}, t_{12} = - \frac{m l n ( 12 )}{b} \dots (ii)

Here, ln

(12) = - 0.693

b = 40 g s^{- 1}, m = 200 g

Substituting in Eq.

(ii)

, we get

t_{1/2} = \frac{0.693 \times 200 g}{40 g s ^{- 1}} = 3.5 s

$2.5 s$

$3.5 s$

$4.5 s$

$7.5 s$

2.5s

3.5s

4.5s

7.5s

$2.5 s$

$3.5 s$

$4.5 s$

$7.5 s$