106v=-kx
==> dv/dt = -kdx/dt
==> a=-kv = -k2x
so it is SHM....
The object moves on the x axis in such a way that its velocity and displacement from the origin satisfy v=-kx , where k is a positive constant
A) The object executes simple harmonic motion
B) The object does not change its direction
C)The kinetic energy of the object keeps on decreasing
D)THe object can change its direction only once
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10 Answers
11i dont think im making any mistake with the question but still
check out jee 07 paper 2 22nd q
11YEs Subhash is agree with you.
v = -kx
Therefore it only depends on the starting position.
If the starting position is positive The velocity is opposite to it.
Therefore The particle goes towards origin with decreasing velocity since the velocity is decreasing as x is decresing.
Therfore the direction is not changed and the kinectic energy reduces as time passes
The ans is B,C
Pink pls
11See what assih did it
v = -kx
dv/dt = -kdx/dt
It is wrong since here kx both are constant with respect to t therefore they cannot be differentiated
1dx/dt = -kx
∫1/x dx = -kt
ln x = -kt +c
x = Ae-kt
v = -kAe-kt
so u could see that the sign of v is constant from t = 0 to t =∞ ... it does not change direction .. B
KE = .5mv2 = 0.5mk2A2e-2kt
decreasing with increasing t ..... C
F ≠-kx .. so no SHM
no change in direction of velocity ...