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riadik2000 [5.3K]

3 years ago

13

A 17.6-kg object oscillates at the end of a horizontal spring that has a spring constant of 20.0 kN/m. The effect of air resista

nce is represented by the damping coefficient b = 6.69 kg/s. Find the time for the amplitude of the oscillation to drop to 5.98% of its original value.

1 answer:

avanturin [10]3 years ago

8 0

Answer:

The time for the amplitude of the oscillation is 14.82 sec.

Explanation:

Given that,

Mass of object = 17.6 kg

Spring constant k= 20.0 kN/m

Damping coefficient b= 6.69 kg/s

Amplitude A= 5.98%

We need to calculate the time for the amplitude of the oscillation

Amplitude at any time is given as

$A_{t}=A_{0}e^{\dfrac{-bt}{2m}}$

Put the value into the formula

$0.0598 A_{0}=A_{0}e^{\dfrac{-6.69\times t}{2\times17.6}}$

$0.0598=e^{\dfrac{-6.69\times t}{2\times17.6}}$

$t=14.82\ sec$

Hence, The time for the amplitude of the oscillation is 14.82 sec.

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