A mass spring system placed on a smooth horizontal surface is oscillating with amplitude xo .At what displacement from the mean position its kinetic energy is twice to that of its potential energy?
Answers
Answer:
Let x be the distance where KE is equal to PE.
Kinetic energy of particles executing SHM is KE=
2
1
mω
2
(a
2
−x
2
)
Potential energy of particles is given by PE=
2
1
mω
2
x
2
Le x be the distance where
KE=PE
2
1
mω
2
(a
2
−x
2
)=
2
1
mω
2
x
2
(a
2
−x
2
)=x
2
x=
2
a
kinetic energy equal to its potential energy at x=
2
a
.
(2) velocity of particles executing SHM is given by v
′
=ω
(a
2
−x
2
)
Maximum velocity v
max
=aω
Let x be the point where the speed becomes half of the maximum speed. Then
v
′
=
2
v
max
ω
(a
2
−x
2
)
=
2
aω
ω
2
(a
2
−x
2
)=
4
a
2
ω
2
a
2
ω
2
−x
2
ω
2
=
4
a
2
ω
2
4a
2
ω
2
−4x
2
ω
2
=a
2
ω
2
3a
2
ω
2
=4x
2
ω
2
x
2
=
4
3a
2
x=
2
3
a
Speed becomes half of the maximum speed at x=
√3a/2
Answer Xo/√3
Explanation:its answer is Xo/√3
Calculations are in the picture
