A pendulum with a mass of 250 g is released from its maximum displacement of 5 degrees and swings with a simple harmonic motion. If the period is 4 seconds, calculate a) the length of the pendulum, b) the amplitude, c) frequency, d) maximum velocity, e) total energy, f) maximum height.

(a) The length of the pendulum is 3.96 m.

(b) The amplitude is 7.89 m.

(c) The frequency is 0.25 Hz.

(d) The maximum velocity is 12.38 m/s.

(e) The total energy is 9.67 J.

(f) The maximum height is 3.94 m.

The length of the pendulum at the given parameters is determined using the following formula.

[tex]T = 2\pi \sqrt{\frac{l}{gcos(\theta)} } \\\\\frac{T}{2\pi } = \sqrt{\frac{l}{gcos(\theta)} }\\\\\frac{T^2}{4\pi ^2 } =\frac{l}{gcos(\theta)}\\\\l = \frac{gcos(\theta)T^2}{4\pi^2} \\\\l = \frac{(9.8)cos(5)\times (4)^2}{4\pi^2} \\\\l = 3.96 \ m[/tex]

The amplitude of the pendulum is calculated as follows;

mgLcosθ = ¹/₂FA

mgLcosθ = ¹/₂(mg)A

Lcosθ = ¹/₂A

A = 2Lcosθ

A = 2 x 3.96 x cos(5)

A = 7.89 m

[tex]\omega = \sqrt{\frac{g}{l} } \\\\\omega = \sqrt{\frac{9.8}{3.96} } \\\\\omega = 1.57 \ rad/s[/tex]

ω = 2πf

f = ω/2π

f = (1.57) / (2π)

f = 0.25 Hz

The maximum velocity is calculated as follows;

v = Aω

v = 7.89 x 1.57

v = 12.38 m/s

E = mgLcosθ

E = 0.25 x 9.8 x 3.96 x cos(5)

E = 9.67 J

h = Lcosθ

h = 3.96 x cos(5)

h = 3.94 m

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