The block shown in the figure is acted on by a sping with spring constant (k) and (a) weak frictional force of constant of constant magitude (f.) The block is pulled a distance `x_(0)` from equilibrium position and then then then fren released. It oscillates many times ultimately comes to rest.

(a) Show that the decrease of amplitude is the same for each cycle of lscillation.
(b) Find the number of cycles the mass oscillates before coming to rest.

Answer

A, B, D

Explanation

(a) From energy conservation principle,
Work done against friction=dicrease in elastic PE
or `f(x_(0)+a_(1)) =1/2k(x_(0)^(2)-a_(0)^(2))`
or `x_(0)=a_(1) =(2f)/k`...(i)
From Eq. (i), we see that decrease of amplitude
`(x_(0)-a_(1))` is `(2f)/k`, which is constant and same for each cycle of oscillation.

(b) The block will come to rest when `ka=f`
`a=k/f`
In the similar manner, we can write
`a_(1) -a_(2)=(2f)/k`....(i)
`a_(2) -a_(3) =(2f)/k`......(ii)
`a_(n-1) -a_(n) =(2f)/k....(n)`
Adding Eqs. `(i), (ii)`,...etc., we get
`x_(0)-a_(n) =n((2f)/k)`
or `a_(n) =x_(0)-n((2f)/k)`.....(B)
Equating Eq. (A) and (B), we get
`k/f =x_(0)-n((2f)/k)`
or `n=(x_(0)-k/f)/(2f) = (kx_(0))/(2f)-(1)/(2)`
Number of cycles,
`m=n/2(kx_(0))/(4f)-(1)/(4)=(1)/(4)((kx_(0))/(f)-1)`.