-->

Mechanical Properties Of Fluids

Question
CBSEENPH11019145

If displacement of a particle at any instant is represented by y = A cos ωt + B sin ωt, where A and B are constants, prove that the motion of particle is simple harmonic motion. What is the amplitude and epoch of the motion?

Solution

Position of particle is given by, 
y = A = cosomega t + B sinomega t               ...(1)
The velocity of particle is given by, 
 straight v space equals fraction numerator d y over denominator d t end fraction equals fraction numerator straight d over denominator d t end fraction left parenthesis straight A space c o s space omega t space plus space straight B space s i n space w t right parenthesis
space space space equals space minus space A omega space s i n space omega t plus B omega space c o s space omega t right parenthesis 
Acceleration of particle is given by,
straight a equals dv over dt equals straight d over dt left parenthesis negative Aω space sin space ωt plus space Bω squared space sin space ωt right parenthesis

space space space equals space minus space Aω squared cosωt minus Bω squared sin space ωt

space space space equals negative straight omega squared left parenthesis Acos space ωt plus straight B space sin space ωt right parenthesis

space space space equals negative straight omega squared straight y
Since the acceleration of particle is directly proportional to displacement and directed towards mean position, therefore the motion is simple harmonic motion.
Now, 
Let space amplitude comma space straight A equals straight r space sinϕ space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis

and space space space space space space space space space space space straight B space equals space straight r space cosϕ space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 3 right parenthesis space

Substituting space space straight A space and space straight B space in space left parenthesis 1 right parenthesis comma space we space get comma

straight y space equals space straight r space sin space straight ϕ space cos space ωt space plus space straight r space cos space straight ϕ space space sin space ωt

space space space equals straight r left parenthesis cos space ωt space sin space straight ϕ plus sin space ωt space cosϕ right parenthesis

space space space equals space rsin left parenthesis ωt plus straight ϕ right parenthesis
Squaring (2) and (3) and adding, we have
space space space space space space straight A squared plus straight B squared equals straight r squared

rightwards double arrow straight r square root of straight A squared plus end root straight B squared  
Dividing (2) (3), we have
space space space space space space fraction numerator space straight A over denominator straight B end fraction equals tan space straight ϕ space space
rightwards double arrow space space space space space straight r equals square root of straight A squared plus straight B squared end root 
Therefore comma space
space A m p l i t u d e space equals space straight r equals space square root of straight A squared plus straight B squared end root
Epoch = straight ϕ space equals space tan to the power of negative 1 end exponent left parenthesis straight A over straight B right parenthesis.