 Physics
6 September, 04:52

# A damped harmonic oscillator loses 6.0% of its mechanical energy per cycle.a. by what percentage does its frequency (equation 14-20) differ from its natural frequency?b. after how many periods will the amplitude have decreased to 1/2 of its original value?

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Answers (1)
1. 6 September, 06:02
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1. The problem statement, all variables and given/known data

A damped harmonic oscillator loses 6.0% of it's mechanical energy per cycle. (a) By what percentage does it's frequency differ from the natural frequency f

0

=

(

1

2

π

)

k

m

? (b) After how many periods will the amplitude have decreased to

1

e

of it's original value?

2. Relevant equations

natural frequency

f

0

=

(

1

2

π

)

k

m

damped frequency

f

=

1

2

π

k

m

-

b

2

4

m

2

displacement for lightly damped harmonic oscillator

x

=

A

e

(

-

b

2

m

)

t

c

o

s

ω

t

Total mechanical energy

E

=

1

2

k

A

2

=

1

2

m

v

2

m

a

x

And I know the mean half life,

2

m

b

is the time until oscillations reach 1/e of original.

3. The attempt at a solution

I used the A^2 expression for E and the A decay term,

A

e

(

-

b

2

m

)

t

, said it loses 6% of E when A^2 =.94A^2 (original) or in other words when

A

e

(

-

b

2

m

)

t

=

0.94

A

so,

e

(

-

b

2

m

)

t

=

.94

-

b

2

m

t

=

1

2

ln (.94)

t =

- 1. The problem statement, all variables and given/known data

A damped harmonic oscillator loses 6.0% of it's mechanical energy per cycle. (a) By what percentage does it's frequency differ from the natural frequency f

0

=

(

1

2

π

)

k

m

? (b) After how many periods will the amplitude have decreased to

1

e

of it's original value?

2. Relevant equations

natural frequency

f

0

=

(

1

2

π

)

k

m

damped frequency

f

=

1

2

π

k

m

-

b

2

4

m

2

displacement for lightly damped harmonic oscillator

x

=

A

e

(

-

b

2

m

)

t

c

o

s

ω

t

Total mechanical energy

E

=

1

2

k

A

2

=

1

2

m

v

2

m

a

x

And I know the mean half life,

2

m

b

is the time until oscillations reach 1/e of original.

3. The attempt at a solution

I used the A^2 expression for E and the A decay term,

A

e

(

-

b

2

m

)

t

, said it loses 6% of E when A^2 =.94A^2 (original) or in other words when

A

e

(

-

b

2

m

)

t

=

0.94

A

so,

e

(

-

b

2

m

)

t

=

.94

-

b

2

m

t

=

1

2

ln (.94)

t =

-

m

b

ln (.94)

But this is time and I need it to be one cycle so do I plug the period in for t?

T = 1/f or 2∏ ω?

m

b

ln (.94)

But this is time and I need it to be one cycle so do I plug the period in for t?

T = 1/f or 2∏ ω?
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