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Question
Answer in brief:
Derive an expression for the period of motion of a simple pendulum. On which factors does it depend?
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Solution
a) Consider a simple pendulum of mass ‘m’ and length ‘L’.
L = l + r,
where l = length of string
r = radius of bob
b) Let OA be the initial position of pendulum and OB, its instantaneous position when the
the string makes an angle θ with the vertical.
In the displaced position, two forces are acting on the bob:
- Gravitational force (weight) ‘mg’ in a downward direction
- Tension T′ in the string.
c) Weight ‘mg’ can be resolved into two rectangular components:
- Radial component mg cos θ along OB and
- Tangential component mg sin θ perpendicular to OB and directed towards mean
position.
d) mg cos θ is balanced by tension T′ in the string, while mg sin θ provides restoring force
∴ F = − mg sin θ
where a negative sign shows that force and angular displacement are oppositely directed.
Hence, restoring force is proportional to sin θ instead of θ. So, the resulting motion is not
S.H.M.
e) If θ is very small then,
sin θ ≈ θ = `"x"/"L"`
∴ F = `- "mg" "x"/"L"`
∴ `"F"/"m" = - "g" "x"/"L"`
∴ `"ma"/"m" = - "g""x"/"L"`
∴ "a" = - "g"/"L" "x" ....(i)
∴ a α - x ....`[therefore "g"/"L" = "constant"]`
f) In S.H.M,
a = − ω2 x ….(ii)
Comparing equations (i) and (ii), we get,
`ω^2 = "g"/"L"`
But, ω = `(2pi)/"T"`
`therefore ((2pi)/"T")^2 = "g"/"L"`
`therefore (2pi)/"T" = sqrt("g"/"L")`
`therefore "T" = 2pi sqrt("L"/"g")` ....(iii)
Equation (iii) represents time period of simple pendulum.
g) Thus period of a simple pendulum depends on the length of the pendulum and
acceleration due to gravity.
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