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(`\frac {Ï€}{8}`) t

What is the displacement of the object after 2s? 

(Ans: 0.25 m, `\frac {1}{16}`Hz, 16 s, x = 0.18 m )



Data Given:

Equation 

x = 0.25 cos(`\frac {Ï€}{8}`) t


To Find:


Amplitude  = x₀ = ?

Frequency  = f = ?

Time period  = T = ?

displacement  = x = ? {after t = 2 s}


Solution:


The instantaneous distance general formula is

x = x₀ cos à´§ t

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

x = x₀ cos 2 Ï€ f t  --------------(1)

and given equation

x = 0.25 cos(`\frac {Ï€}{8}`) t  ------------(2)

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


Amplitude x₀ :

x₀ = 0.25 m ---------------Ans. (1)


Frequency f:

2 Ï€ f t = (`\frac {Ï€}{8}`) t

or
f = `\frac {1}{16}` Hz  --------------Ans. (2) 


Time period T:

as we know that 

T = `\frac {1}{f}` = 16 s  --------------Ans.(3)


Dispalcement x at t = 2 s :

By putting t = 2 s in equation (2)

x = 0.25 cos ((`\frac {Ï€}{8}`) 2 )

x = 0.25 cos(`\frac {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)



************************************

Shortcut Links For 


1. Website for School and College Level Physics   
2. Website for School and College Level Mathematics  
3. Website for Single National Curriculum Pakistan - All Subjects Notes 

© 2022-Onwards by Academic Skills and Knowledge (ASK)     

Note:  Write me in the comment box below for any query and also Share this information with your class-fellows and friends.