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Lana71 [14]
3 years ago
11

A damped oscillator is released from rest with an initial displacement of 10.00 cm. At the end of the first complete oscillation

, the displacement reaches 9.05 cm. When 4 more oscillations are completed, what is the displacement reached
Physics
1 answer:
Tpy6a [65]3 years ago
5 0

Answer:

The  displacement is  A_r =  6.071 \ cm

Explanation:

From the question we are told that

   The initial displacement is A_o  =  10 \ cm

     The displacement at the end of first oscillation is  A_d  = 9.05 \ cm

     

Generally the damping constant of this damped oscillator is mathematically represented as  

           \eta  =  \frac{A_d}{A_o}

substituting values

           \eta  =  \frac{9.05}{10}

        \eta  =  0.905

The displacement after 4 more oscillation is mathematically represented as

       A_r =  \eta^4 *  A_d

substituting values

      A_r =  (0.905)^4 *   (9.05)

      A_r =  6.071 \ cm

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<u> </u><u>»</u><u> </u><u>Image</u><u> </u><u>distance</u><u> </u><u>:</u>

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

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{ \tt{ \frac{1}{v} +  \frac{1}{10} =  \frac{1}{5}   }} \\  \\  { \tt{ \frac{1}{v}  =  \frac{1}{10} }} \\  \\ { \tt{v = 10}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \:  \: }}}}}

<u> </u><u>»</u><u> </u><u>Magnification</u><u> </u><u>:</u>

• Let's derive this formula from the lens formula:

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

» Multiply throughout by fv

{ \tt{fv( \frac{1}{v} +  \frac{1}{u} ) = fv( \frac{1}{f}  )}} \\   \\ { \tt{ \frac{fv}{v}  +  \frac{fv}{u}  =  \frac{fv}{f} }} \\  \\  { \tt{f + f( \frac{v}{u} ) = v}}

• But we know that, v/u is M

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{ \tt{M =  \frac{10}{5} - 1 }} \\  \\ { \tt{M = 5 - 1}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}

<u> </u><u>»</u><u> </u><u>Nature</u><u> </u><u>of</u><u> </u><u>Image</u><u> </u><u>:</u>

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