A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (a) 5 cm (b) 3 cm (c) 0 cm.

Explanation

Amplitude, A=5cm=0.05
Time period , T=0.2s
(i) For displacement , x=5cm=0.05m
Accelertion is given by : `a=-omega^(2)x` Velocity is given by:
`v=omegasqrt(A^(2)-x^(2))=(2pi)/(T)sqrt((0.05)^(2)-(0.05)^(2))=0`
When the displacement of the body is 5cm, its accelertion is `-5pi^(2)m//s^(2)` and velocity is 0.
(ii)For displacement , x= 3cm=0.03m
`a=-omega^(2)xx=-((2pi)/(T))^(2)xx=-((2pi)/(0.2))^(2)xx0.03=-3pi^(2)m//s^(2)`
Accelertion is given by:
Velocity is given by: `v=omegasqrt (A^(2)-x^(2))`
`v=(2pi)/(T)sqrt(A^(2)-x^(2))=(2pi)/(T)sqrt((0.05)^(2)-(0.03)^(2))=(2pi)/(0.2)xx0.04=0.4pim//s`
When the displacement of the body is 3cm , its accelertion is `-3pim//s^(2)` and velocity is `0.4pim//s`
(ii) For displacement ,x=0
Acceleration is given by: `a=-omega^(2)xx=0`
`v=omegasqrt(A^(2)-x^(2))=(2pi)/(T)sqrt(A^(2)-x^(2))`
Velocity is given by:
=`(2pi)/(0.2)sqrt((0.05)^(2)-0)=0.5pim//s`
When the displacement of the body is 0,its accelertion is 0and velocity `0.5pim//s`.