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Question
A mass of 2 kg is attached to the spring of spring constant 50 Nm–1. The block is pulled to a distance of 5 cm from its equilibrium position at x = 0 on a horizontal frictionless surface from rest at t = 0. Write the expression for its displacement at anytime t.
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Solution
Consider the diagram of the spring block system. It is an S.H.M. with an amplitude of 5 cm about the mean position shown.
Given, the spring constant k = 50 N/m
m = mass attached = 2 kg
∴ Angular frequency `ω = sqrt(k/m)`
= `sqrt(50/2)`
= `sqrt(25)`
= 5 rad/s
Assuming the displacement function
`y(t) = Asin(ωt + phi)`
Where `phi` = initial phase
But given at t = 0, y(t) = + A
y(0) = + A = Asin(ω × 0 + `phi`)
or `sin phi` = 1 ⇒ `phi = pi/2`
∴ The desired equation is `y(t) = Asin(ωt + pi/2) = Acos ωt`
Putting A = 5 cm, ω = 5 rad/s
We get, y(t) = 5sin5t
Where t is in second and y is in centimetre.
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