Question
A body executing linear simple harmonic motion has a velocity of $$3\,m/s$$ when its displacement is $$4\,cm$$ and a velocity of $$4\,m/s$$ when its displacement is $$3\,cm.$$ What is the amplitude of oscillation?
A.
$$5\,cm$$
B.
$$7.5\,cm$$
C.
$$10\,cm$$
D.
$$12.5\,cm$$
Answer :
$$5\,cm$$
Solution :
Velocity in $$SHM$$ is given by
$$\eqalign{
& v = \omega \sqrt {{a^2} - {y^2}} \cr
& {\text{At}}\,\,y = 4\,cm = 0.04\,m,v = 3\,m/s \cr
& \therefore 3 = \omega \sqrt {{a^2} - {{\left( {0.04} \right)}^2}} \,......\left( 1 \right) \cr
& {\text{At}}\,\,y = 3\,cm = 0.03\,m,v = 4\,m/s \cr
& \therefore 4 = \omega \sqrt {{a^2} - {{\left( {0.03} \right)}^2}} \,......\left( 2 \right) \cr} $$
Dividing (2) by (1),
we get $$a = 0.05 = 5\,cm$$