Amplitude of a mass-spring system, which is executing simple harm… - Vidyadip

MCQ

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 ...... $s$ time is required for the amplitude of the system to drop to half of its initial value ? $(\ln 2=0.693)$

✓
$34.65$
B
$17.32$
C
$0.034$
D
$15.01$

✓

Answer

Correct option: A.

$34.65$

a
$A = A _{0} e ^{-\gamma t }= A _{0} e ^{-\frac{ bt }{2 m }}$

$\frac{ A _{0}}{2}= A _{0} e ^{-\frac{ bt }{2 m }}$

$\frac{ bt }{2 m }=\ln 2$

$t =\frac{2 m }{ b } \ln 2=\frac{2 \times 500 \times 0.693}{20}$

$t =34.65\, second.$

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