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BITSAT 2019 Paper

Section: Physics

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Question : 24 of 150

Marks: +1, -0

The amplitude of a damped oscillator becomes

\left(\frac{1}{3}\right)^{\text{rd}}

in 2 seconds. If its amplitude after 6 seconds is

\left(\frac{1}{n}\right)

times the original amplitude, the value of n is

$3^2$
$3^3$
$\sqrt[3]{3}$
2 ${}^3$

Solution:

(b) Amplitude of a damped oscillator

A =

A_0 e^{\frac{-bt}{2m}}

Case 1:-

When t = 2 s, A =

\frac{A_0}{3}

∴

\frac{A_0}{3}

=

A_0 e^{\frac{-2b}{2m}}

⇒

\frac{1}{3}

=

e^{\frac{-b}{m}}

..........(i)

Case 2:-

When t = 6s, A =

\frac{A_0}{n}

∴

\frac{A_0}{n}

=

A_0 e^{\frac{-6b}{2m}}

⇒

\frac{1}{n}

=

\left(e^{\frac{-b}{m}}\right)^3

..........(ii)

From (i) and (ii)

\frac{1}{n} = \left(\frac{1}{3}\right)^3

⇒ ∴ n =

3^3

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