Amplitude of a mass spring system, which is executing simple harmonic motion decreases with time. If $m a s s = 500 g$ , $D e c a y c o n s \tan t = 20 g / s$ then how much time is required for the amplitude of the system to drop to half of its initial value?

$15.01 s$

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$17.32 s$

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$0.034 s$

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$34.65 s$

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Solution

The correct option is D
$34.65 s$

Step 1: Given Data:
Mass, $m = 500 g$
Damping constant, $b = 20 \frac{g}{s}$
Initial amplitude $= A_{0}$
Final amplitude, $A = \frac{A_{0}}{2}$
$\ln 2 = 0.693$
Step 2: Formula Used:
$A = A_{0} e^{\frac{- b t}{2 m}}$
Where, $A =$ Amplitude of oscillation
$b =$ Decay constant
$m =$ mass
$t =$ time
Step 3: Calculating the time:
$A = A_{0} e^{\frac{- b t}{2 m}}$
$\frac{A_{0}}{2} = A_{0} e^{\frac{- 20 \times t}{2 \times 500}}$
$\frac{1}{2} = e^{\frac{- t}{50}}$
$\frac{t}{50} = \ln 2$
$t = 50 \times 0.693$
$= 34.65 s$
Hence, option D is the correct answer.

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

Amplitude of a mass spring system, which is executing simple harmonic motion decreases with time. If $m a s s = 500 g$ , $D e c a y c o n s \tan t = 20 g / s$ then how much time is required for the amplitude of the system to drop to half of its initial value?