Question

A small earthquake starts a lamppost vibrating back and forth. The amplitude of the vibration of the top of the lamppost is 6.4 cm at the moment the quake stops, and 8.8 s later it is 1.6 cm .What is the time constant for the damping of the oscillation? What was the amplitude of the oscillation 4.4 s after the quake stopped?

Answer 1

Answer:

$\tau = 6.35 s$

A = 3.2 cm

Explanation:

given,

Amplitude of the vibration of the top of lamppost = 6.4 cm

after 8.8 s the amplitude is 1.6 cm

time constant for damping of oscillation = ?

Amplitude at 4.4 second= ?

using formula

$A = A_0e^{-\dfrac{T}{\tau}}$

$1.6 = 6.4\times e^{-\dfrac{8.8}{\tau}}$

taking ln both side

$ln (1.6) = ln(6.4)-\dfrac{8.8}{\tau}$

$\tau = 6.35 s$

$A = A_0e^{-\dfrac{T}{\tau}}$

$A =6.4\times e^{-\dfrac{4.4}{6.35}}$

A = 3.2 cm

Answer 2

The amplitude of the oscillation 4.4 s after the quake stopped is:

A= 3.2cm

What is Amplitude?

This refers to the maximum length which an object can attain when it oscillates or vibrates.

Given that the amplitude from the top of the lamppost is given as:

6.4cm

We need to find the time constant for the damping of the oscillation.

We use the formula

$A= Aoe$^-T/t

We would input the values

1.6 = 6.4 x $e$^-8.8/t

Taking Ln on both sides

Ln(1.6) = ln(6.4)^-8.8/t

t=6.35₈

A= $Aoe$^T/t

A= 3.2cm

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