A particle executing S.H.M. has velocity 10 cm/s and 8 cm/s at distance 4 vm&5 cm respectively. The period of oscillation of the particle, in seconds,is
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Answer:
Velocity of particle executing SHM at 2 cm from mean position v
1
=4 cm/s and at 3 cm from mean position v
2
=3 cm/s.
v
1
=ω
a
2
−x
2
=ω
a
2
−4
v
2
=ω
a
2
−x
2
=ω
a
2
−9
v
1
v
2
=
a
2
−4
a
2
−9
=
4
3
16a
2
−144=9a
2
−36
a
2
=
7
108
Putting the value of a in first equation we get :
ω=1.18s
−1
Time period, T=
ω
2π
=5.3 s
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