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Question
A particle performing S.H.M. has velocities of 8 cm/s and 6 cm/s at displacements of 3 cm and 4 cm respectively. Calculate the amplitude and period of S.H.M.
Solution
Given:
v1 = 8 cm/s, v2 = 6 cm/s, x1 = 3 cm, x2 = 4 cm
To find: Amplitude (A), Period (T)
Formula: v = ω`sqrt("A"^2 - "x"^2)`
Calculation:
From the given condition,
`"v"_1^2 = ω^2("A"^2 - "x"_1^2)`
∴ 64 = ω2(A2 − 9) .…(i)
Also, `"v"_2^2 = ω^2("A"^2 - "x"_2^2)`
∴ 36 = ω2(A2 − 16) …(ii)
Dividing (i) by (ii),
`64/36 = ("A"^2 - 9)/("A"^2 - 16)`
∴ `16/9 = ("A"^2 - 9)/("A"^2 - 16)`
∴ 16A2 − 256 = 9A2 − 81
∴ 7A2 = 175
∴ A = 5 cm
Substituting value of A in equation (i), we get,
64 = ω2(25 − 9) = 16 ω2
∴ ω2 = 4
∴ ω = 2 rad/s
∴ `(2pi)/"T" = 2`
∴ T = π s = 3.14 s
The amplitude and period of S.H.M. of the particle are 5 cm and 3.14 s respectively.
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