A pendulum with time period of 1 s is losing energy due to damping. At certain time its energy is 45 J. If after completing 15 oscillations, its energy has become 15 J, its damping constant (in s-1) is :

Question

A pendulum with time period of 1 s is losing energy due to damping. At certain time its energy is 45 J. If after completing 15 oscillations, its energy has become 15 J, its damping constant (in s-1) is : (A) (1/30) ln3 (B) (1/15) ln3 (C) 2 (D) (1/2)

Tardigrade Admin · Accepted Answer

Amplitude of a damped oscillator A=A o e-k t where b is the damping constant. Energy E=(1/2) K A2 E=((1/2) K A02) e-2 bt ⇒ 15=45 e-2b( 1× 15) 3=e30 b ⇒ b=(1/30) ln (3)

Q. A pendulum with time period of $1 s$ is losing energy due to damping. At certain time its energy is $45 J$ . If after completing $15$ oscillations, its energy has become $15 J$ , its damping constant (in $s^{- 1}$ ) is :

$\frac{1}{30} ln 3$

$\frac{1}{15} ln 3$

$2$

$\frac{1}{2}$

Amplitude of a damped oscillator
$A = A o e^{- k t}$
where $b$ is the damping constant.
Energy $E = \frac{1}{2} K A^{2} E = (\frac{1}{2} K A_{0}^{2}) e^{- 2 b t}$
$\Rightarrow 15 = 45 e^{- 2 b (1 \times 15)} 3 = e^{30 b}$
$\Rightarrow b = \frac{1}{30} ln (3)$

Q. A pendulum with time period of 1s is losing energy due to damping. At certain time its energy is 45J. If after completing 15 oscillations, its energy has become 15J, its damping constant (in s−1) is :

301​ln3

151​ln3

2

21​

Amplitude of a damped oscillator A=Aoe−kt where b is the damping constant. Energy E=21​KA2E=(21​KA02​)e−2bt ⇒15=45e−2b(1×15)3=e30b ⇒b=301​ln(3)

Q. A pendulum with time period of $1 s$ is losing energy due to damping. At certain time its energy is $45 J$ . If after completing $15$ oscillations, its energy has become $15 J$ , its damping constant (in $s^{- 1}$ ) is :

$\frac{1}{30} ln 3$

$\frac{1}{15} ln 3$

$2$

$\frac{1}{2}$

Amplitude of a damped oscillator
$A = A o e^{- k t}$
where $b$ is the damping constant.
Energy $E = \frac{1}{2} K A^{2} E = (\frac{1}{2} K A_{0}^{2}) e^{- 2 b t}$
$\Rightarrow 15 = 45 e^{- 2 b (1 \times 15)} 3 = e^{30 b}$
$\Rightarrow b = \frac{1}{30} ln (3)$