A simple pendulum consists of a ball connected to one end of a thin brass wire. The period of the pendulum is 2.51 s. The temperature rises by 142 C°, and the length of the wire increases. Determine the change in the period of the heated pendulum.

To find the change in period of the heated pendulum, we first need to determine the length of the pendulum using the formula for the period of the pendulum, T = 2π√(L/g) where L = length of pendulum and g = acceleration due to gravity = 9.8 m/s². Making L subject of the formula, we have

L = gT²/4π²

Since T = 2.51 s, substituting the values of the variables into the equation, we have

L = gT²/4π²

L = 9.8 m/s² × (2.51)²/4π²

L = 9.8 m/s² × 6.3001 s²/4π²

L = 64.741 m/39.478

L = 1.64 m

Since the temperature of the thin brass rises by 142 °C, its change in length ΔL = LαΔθ where L = length of thin brass = 1.64 m, α = linear expansivity of brass = 2 × 10⁻⁵ /°C and Δθ = temperature rise = 142 °C

So, substituting the values of the variables into the equation, we have

ΔL = LαΔθ

ΔL = 1.64 m × 2 × 10⁻⁵ /°C × 142 °C

ΔL = 465.76 × 10⁻⁵ m

ΔL = 0.0046576 m

To find the change in period of the heated pendulum, we differentiate T with respect to L to get

dT/dL = d(2π√(L/g))/dL

dT/dL = (2π/√g)d(√L)/dL

dT/dL = π/√(gL)

So, ΔT = dT/dL × ΔL

ΔT = π/√(gL) × ΔL

Substituting the values of the variables into the equation, we have

ΔT = π/√(gL) × ΔL

ΔT = π/√(9.8 m/s² × 1.64 m) × 465.76 × 10⁻⁵ m

ΔT = π/√(16.072 m²/s²) × 465.76 × 10⁻⁵ m

ΔT = π/4.0089 m/s) × 465.76 × 10⁻⁵ m

ΔT = 0.0786 s/m × 465.76 × 10⁻⁵ m

ΔT = 364.99 × 10⁻⁵ s

ΔT = 0.0036499 s

ΔT ≅ 0.00365 s

So, the change in period of the heated pendulum is 0.00365 s

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