Find the amplitude, frequency and period of an object vibrating at the end of a spring, if the equation for its position, as a function of time, is:

x = 0.25 cos(Ï€8) t

What is the displacement of the object after 2s? 

(Ans: 0.25 m, 116Hz, 16 s, x = 0.18 m )



Data Given:

Equation 

x = 0.25 cos(Ï€8) t


To Find:


Amplitude  = xâ‚€ = ?

Frequency  = f = ?

Time period  = T = ?

displacement  = x = ? {after t = 2 s}


Solution:


The instantaneous distance general formula is

x = xâ‚€ cos à´§ t

where à´§ = 2Ï€T =  2 Ï€ f   {as f = 1/T} so 

x = xâ‚€ cos 2 Ï€ f t  --------------(1)

and given equation

x = 0.25 cos(Ï€8) t  ------------(2)

On comparing equation (1) and (2) we get


Amplitude xâ‚€ :

xâ‚€ = 0.25 m ---------------Ans. (1)


Frequency f:

2 Ï€ f t = (Ï€8) t

or
f = 116 Hz  --------------Ans. (2) 


Time period T:

as we know that 

T = 1f = 16 s  --------------Ans.(3)


Dispalcement x at t = 2 s :

By putting t = 2 s in equation (2)

x = 0.25 cos ((Ï€8) 2 )

x = 0.25 cos(Ï€4)  

x = 0.25 cos 45°

x = 0.25 ｘ 0.707 m

x = 0.17675 m

or

x = 0.18 m  -----------------Ans. (4)



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