1) A student says that he had applied a force F = - k√x on a particle and the particle moved in SHM. He refuses to tell whether k is constant or not. assume that he has worked only with positive x axis and no other forces acts on the particle.
a) As x increases k increases ( this is the correct answer)
b)The motion cannot be simple harmonic ( I think this should also be the correct answer because for any particle to perform SHM F= - kx { a must condition}, but here F = -k√x. I am confused here so please correct me)
2)A particle executes SHM with frequency ν. The frequency with which the kinetic energy oscillates
a)ν/2
b)ν
c)2ν
d)zero
112)SHM is a kind of oscillation.
thus frequency of oscillation is v
kinetic energy oscillation is 2v!
How to get this answer, means is there any way to solve it or any formula to get the answer?
You have only given the answer directly, is there any reason for it to be true.
11@ Vivek, really are you ?Είμαι σε καθαÏή και ο χÏÏ
σός ( I am pure as gold) [3]
71Είμαι σε καθαÏή και ο χÏÏ
σός !
Nice. :P
49x=Asin(wt+θ)
v=Awcos(wt+θ)
KE= 12mv2=12mA2w2cos2(wt+θ)=12mA2w21+cos2(wt+θ)2=14mA2w2[1+cos2(wt+θ)]=14mA2w214mA2w2cos2(wt+θ)
thus, frequency = 2w
11@ Vivek , my net was not working properly, it was getting disconnect every noe and then so jyada time laga.
49@Khyati:
Look....
in SHM the constant m in the equation F=-mx is constant
but in the given thing, F=-k√x
k may or may not be constant
if k is constant, then the motion CANNOT be SHM
but if the motion has to be SHM then F=-mx where m is constant
comparing the given and assumed formula
F=-mx=-k√x
this gives, if the motion is SHM then k=m√x
now as x increases, m being constant, k must increase as proportional to √x if the motion is to be SHM and in the question it is given that the motion is SHM!
1abt
x=Asinwt
=Asin2pieft
v=2piefAcos2pieft
1/2mv^2=2pie^f^A^2/m cos^2 2pieft
now cos^2 2@=1-sin2@/2
or 1+sin2@/2 not sure but whatever the frequency will be twice
and abt the particle in a circle i think is should be an oscillatory motion but correct me if i am wrong....
an object can only be in oscillatory motion only if it is in periodic motion
71[Woh 10.30 Pm to 01.16 AM] Woh,,,
By the way nice share khyati! Thanks Subho.
113)The free end of the simple pendulum is attached to the ceiling of of a box. The box is taken to the height and the pendulum is oscillated. When the bob is at its lowest point the bob is released to fall freely. As seen from the box during this period, the bob will go in a circular path. WHY? Won't it stop oscillating as time period will tend to infinity because of the free fall?
4)A particle moves in a circular path with a continuously increasing speed. Its motion
a)period
b)oscillatory
c)none of them (correct answer given in the book)
Why option (a) be the correct answer ?
11@ Subhomoy, I have doubt in 1st.
I have read F = -kx is the definition of SHM, so shouldn't all the particles having SHM must satisfy this equation, but in the given question F = -k√x, isn't this equation different than the equation which forms the necessary condition as well as the definition of SHM.
495) well gravitational field inside earth has expression valid for SHM inside the earth where F=-Kx
but in the distance h above the ground, the force is related with inverse square of the distance!!
thus not SHM!
494) in periodic motion, a body must return to each of it's positions after a fixed time interval...
thus, the case in consideration is not periodic as the clause fixed time is not maintained!
493)
no!!
in the lowest position, the particle has non-zero horizontal velocity..
at this moment the force acting on the body (in non inertial frame) becomes zero...
thus in the non inertial frame, it is equivalent to a stone tied to a rope and given tangential velocity in a gravity free chamber....
thus it executes circular motion!!
492) (with nishant bhaiya's assurance)
SHM is a kind of oscillation.
thus frequency of oscillation is v
kinetic energy oscillation is 2v! :P
491) particle moves in SHM (given)
so, F=-mx (m=arbitrary const)
bt given, F=-k√x
comparing we get , k=m√x
satisfying first option!
115)Suppose a tunnel is dug along the diameter of the earth. A particle is dropped from a point, a distance h directly above the tunnel. The motion of the particle as seen from the earth is
a) SHM
b)on a straight line
c)periodic
why can't option (a) also be correct ?