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Dmitrij [34]
3 years ago
14

A spring with a spring constant k of 100 pounds per foot is loaded with 1-pound weight and brought to equilibrium. It is then st

retched an additional 1 inch and released. Find the equation of motion, the amplitude, and the period. Neglect friction.
Physics
1 answer:
sukhopar [10]3 years ago
6 0

Answer:

Explanation:

angular frequency of resulting SHM ω = √(k/m

ω  =√(100x32/1

ω  = 56.57

Period T

ω = 2π /T = 56.57

T = 2π / 56.57

= .11 s .

spring is stretched by 1 inch so it will become the amplitude of frequency

A = 1 inch

when t = 0 , x ( displacement ) = A = 1 inch

equation of motion

x ( t ) in inch  =  A cosω t

= 1 cos 56.57t

x ( t ) in inch  =  1 cos 56.57 t .

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Answer:

Explanation:

Given

length of window h=2.9\ m

time Frame for which rock can be seen is \Delta t=0.134\ s

Suppose h is height above which rock is dropped

Time taken to cover h+2.9 is t_1

so using equation of motion

y=ut+\frac{1}{2}at^2

where  y=displacement

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time taken to travel h  is

h=0+0.5\times g\times (t_2)^2---2

Subtract 1 and 2 we get

2.9=0.5g(t_1^2-t_2^2)

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and from equation t_1-t_2=0.134\ s

so t_1+t_2=\frac{5.8}{9.8\times 0.134}

t_1+t_2=4.416\ s

and t_1=t_2+\Delta t

so t_2+\Delta t+t_2=4.416

2t_2+0.134=4.416

t_2=0.5\times 4.282

t_2=2.141\ s

substitute the value of t_2 in equation 2

h=0.5\times 9.8\times (2.141)^2

h=22.46\ m

                                                     

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where would information on the chemical and physical properties of a specific chemical be located in a laboratory or in the work
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Answer:

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A sound source is moving at 80 m/s toward a stationary listener that is standing in still air (a) Find the wavelength of the sou
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Answer:

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b. observed frequecy, \lambda = 0.7604\lambda_{o}

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speed of sound in air or vacuum, v_{a} = 343 m/s

speed of sound observed, v_{o} = 0 m/s

Solution:

From the relation:

v = \vartheta \lambda        (1)

where

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(b) The observed frequency is given by:

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Using eqn (2) and (3):

\lambda = \frac{334}{1.315} = \frac{1}{1.315}\frac{v_{a}}{\vartheta_{o}}

\lambda = 0.7604\lambda_{o}

4 0
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