A force of different kind

Classical Mechanics Level 5

A body of mass m m is kept on a smooth horizontal surface at x = 0 x=0

A force F = x + sin ⁡ t F = x+\sin{t} starts acting on the body at t = 0 t=0 .

Find the x x -coordinate of body at t = 5 t=5 up to 3 decimal places

Details and Assumptions

  • m = 1 kg m = 1\text{ kg} .

  • At t = 0 t=0 velocity of block is 0.5 m/s 0.5\text{ m/s} towards negative x x -axis.

This question was made by me and Hargun Singh


The answer is 0.4794.

Neelesh Vij by neelesh vij , 21, India

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1 solution

We find general solution to

x ′ ′ = x + s i n ( t ) x'' = x +sin(t) is:

x = θ e t + η e − t − ( s i n ( t ) / 2 ) x= \theta e^{t} + \eta e^{-t} -(sin(t)/2) .

Now at t = 0 , x = 0 t=0, x=0 , we have θ = − η \theta = -\eta .

v = x ′ = θ e t − η e − t − ( c o s ( t ) / 2 ) v= x'= \theta e^{t} - \eta e^{-t} -(cos(t)/2) .

But v t = 0 = − 0.5 v_{t=0} = -0.5 .

Hence θ = η \theta= \eta

Thence θ = η = 0 \theta= \eta= 0 .

Hence x = − s i n ( t ) / 2 x= -sin(t)/2 .

Therefore answer = x t = 5 = 0.4794 x_{t=5} = \boxed{0.4794} ( a p p r o x ) approx)

Note : F = m x ′ ′ F= m x''

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Forgot to change mode of calculator from degree to radian!

Prince Loomba - 4 years, 4 months ago

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