Solved Numerical Problem No. 10 ( Unit-7: Oscillation); HSSC-1 (Class-11); Physics (New Book/KPK Text Board, Peshawar) With Verified Answers

Find the amplitude, frequency and time period of an object oscillating at the end of a spring, if the equation for its position at any instant given by x = 0.25 cos`(\ frac {π}{8})` t . Find the displacement of the object after 2 s. (0.25 m, `\ frac {1}{16}`Hz, 0.18 m)

Given:

The equation for position of an object at any instant oscillating at the end of a spring 

x = 0.25 cos `(\ frac {π}{8})` t

 

To Find:

Amplitude = x₀ = ?

Frequency = f = ?

Time period = T = ?

Displacement  = x = ? (after time t = 2s)

Solution:

As given  

x = 0.25 cos `(\ frac {π}{8})` t ---------------(1)

But generally, a displacement for an object executing Simple Harmonic Motion (SHM) is given by an equation

x = x₀ cos ധt ---------------------(2)

So by comparing the corresponding values of both of equations (1) and (2) we get

x₀ = Amplitude = 0.25 m-------------- Ans (1)

and

ധ = `\frac {π}{8}` ---------------- (3)

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So to find frequency f  we know

ധ = 2πf             ஃ[`\frac {2π}{T}` and f = `\ frac {1}{T}` ]

or

f = `\frac {ധ }{2π}`

using equation (3)

f = `\frac {π/8}{2π}`

OR

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

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ஃTime period and Frequency are related inversely i.e.  [ T = `\ frac {1}{f}` ]

So

T = `\frac {1}{(1/16) Hz}`

or

T = 16 s ----------------Ans. (iii)

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Now to find displacement x after time t = 2s

Putting the corresponding values  in equation (2) i.e.

x = x₀ cos ധt

x = 0.25 m ｘ cos (`\frac {π}{8}` ｘ 2)

x = 0.25 m ｘ cos `\frac {π}{4}`

x = 0.25 m ｘ cos 45° 

x = 0.25 m ｘ 0.707

x = 0.177 m

0r

x = 0.18 m ---------------------Ans.(4)

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