a particle executing simple harmonic motion. If velocity of particle at distance x1 and x2 from the mean position are u1 and u2 respectively then prove that it's time period will T, T =2pi under root (x2)²-(x1)²/(u1)²-(u2)²
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Answer:
Explanation:
Correct option is D)
Using the expressions for velocity for SHM,
v
1
=ω
A
2
−x
1
2
v
2
=ω
A
2
−x
2
2
Squaring both,
v
1
2
=ω
2
(A
2
−x
1
2
) ....(3)
v
2
2
=ω
2
(A
2
−x
2
2
) ....(4)
Subtracting (4) from (3), v
1
2
−v
2
2
=ω
2
(x
2
2
−x
1
2
)
⇒4π
2
(x
2
2
−x
1
2
)=T
2
(v
1
2
−v
2
2
)
⇒T=2π
v
1
2
−v
2
2
x
2
2
−x
1
2
where we have used ω=
T
2π
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