Physics Numerical Problems No. 7.8, Unit-7: Oscillation, HSSC-I (class 11) Physics (Punjab Curriculum Textbook Board, Lahore) With Verified Answers

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 {π}{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 {π}{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 ധ t

where ധ = `\frac {2π}{T}` =  2 π f   {as f = 1/T} so 

x = x₀ cos 2 π f t  --------------(1)

and given equation

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

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

Amplitude x₀ :

x₀ = 0.25 m ---------------Ans. (1)

Frequency f:

2 π f t = (`\frac {π}{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 {π}{8}`) 2 )

x = 0.25 cos(`\frac {π}{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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