IIT JEE , AIEEE Forum - Physics , Mechanics

vibhav1991

A particle is moving with S.H.M in a straight line. When the distance of the particle from the equilibrium position has values x and y the corresponding values of velocities are u and v. Show that the time period of oscillation is given by-----

T=2 [(y²-x²)/(u²-v²)]^1/2

Conjurer

Without loss of generality

we can say this SHM is the spring block SHM.

And we know that total energy of a particle executing SHM is 1/2mw^2a^2

Total mechanical energy = KE + PE

So 1/2 mw^2a^2 = 1/2 w^2mx^2 + 1/2mu^2

and 1/2 mw^2a^2 = 1/2 w^2my^2 + 1/2 mv^2 ( because k=w^2m)

Dividing both equations:

w^2(y^2-x^2) = (u^2-v^2)

w^2 = (u^2 - v^2)/(y^2-x^2)

T= 2pi/w = 2 [(y²-x²)/(u²-v²)]^1/2

tarun_bits

v= w (A2 - x2)root

put x and u
and then y and v

Eliminate a from the two equations and find w

then T= 2pie / w
like this u get it

tarun_bits

ur method is also good conjurer..

i can show the full solution vibhav by my method if u want

suresan

u^2 = w^2 A^2 - w^2 x^2

v^2 = w^2 A^2 - w^2 y^2

Where w angular frequency and A amplitude

v^2 - u^2 = (y^2 - x^2) w^2

w = 2pi/T =sqrt[ (v^2 - u^2)/(y^2 - x^2)]

T = 2Pi X sqrt[(y^2 - x^2) / (v^2 - u^2)

tarun_bits

U = w root of ( a2 ? x2).......1

V = w roof of (a2 ? y2)..........2

From 1

W = u / root of (a2 ? x2 )

Putting in 2?

V = u root of (a2 ? y2) / root of (a2 ? x2)

Squaring?.

V2 / u2 = a2 ? y2 / a2 ? x2

V2(a2 ? x2) = u2 (a2 ? y2)

So
A2 = x2v2 ? y2u2 / v2 ? u2

Now????.
W2 = u2 / a2 ? x2

Putting value of a2

W2 = u2
--------
X2v2 ? y2u2 - x2
-------------
V2 ? u2

W2 = v2 ? u2 / x2 ? y2

W = root of ( u2 ? v2 / y2 ? x2 )

So
T = 2 pie root of (y2 ? x2 / u2 ? v

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