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:
v=ω
A
2
−x
2
v
1
=10cm/sec
v
2
=7cm/sec
x
1
=3cm
x
2
=4cm
v
1
=ω
A
2
−x
1
2
v
2
=ω
A
2
−x
2
2
∴
7
10
=
ω
A
2
−(4)
2
ω
A
2
−(3)
2
49
100
=
A
2
−16
A
2
−9
51A
2
=1600−(9×49)
A=4.76cm
Length of the path is 2A=9.52 cm
Explanation:
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