Radians
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A high-precision sensor measures the output voltage V V\,V of a microscopic resonator as a function of its deflection angle θ\thetaθ (in radians). The sensor output is modeled by the equation:

V(θ)=3cos⁡(2θ)−3cos⁡(2θ)cos⁡(6θ)V(\theta) = 3\cos(2\theta) - 3\cos(2\theta)\cos(6\theta)V(θ)=3cos(2θ)−3cos(2θ)cos(6θ)

a.

Show that for small values of θ\thetaθ, V(θ)≈54θ2V(\theta) \approx 54\theta^2V(θ)≈54θ2.

[4]
b.

The energy E E\,E dissipated during an oscillation cycle is given by E=∫00.1154V(θ) dθ\displaystyle E = \int_{0}^{0.1} \sqrt{\frac{1}{54}V(\theta)} \, d\thetaE=∫00.1​541​V(θ)​dθ. Show that the energy E E\,E can be approximated by E≈2m×5nE \approx 2^m \times 5^nE≈2m×5n, where m m\,m and n n\,n are integers to be determined.

[5]
ci.

Explain why ∫12.612.7θ dθ\int_{12.6}^{12.7} \theta \, d\theta∫12.612.7​θdθ is not a suitable approximation for ∫12.612.7154V(θ) dθ\displaystyle \int_{12.6}^{12.7} \sqrt{\frac{1}{54}V(\theta)} \, d\theta∫12.612.7​541​V(θ)​dθ.

[1]
cii.

Explain how ∫12.612.7154V(θ) dθ\displaystyle \int_{12.6}^{12.7} \sqrt{\frac{1}{54}V(\theta)} \, d\theta∫12.612.7​541​V(θ)​dθ may be approximated by ∫abθ dθ\int_{a}^{b} \theta \, d\theta∫ab​θdθ for suitable values of a a\,a and bbb.

[2]

Radians Questions

Practise Edexcel A Level Maths Radians with exam-style questions for A Level Maths. 52 questions covering 5.1 Radian Measure, 5.2 Arc Length, 5.3 Areas of Sectors and Segments, 5.4 Solving Trigonometric Equations, and 5.5 Small Angle Approximations, matched to the Edexcel A Level Maths (9MA0) specification and written in Paper 1, Paper 2 and Paper 3 style. Every question includes a full worked solution and mark scheme, so you can see where marks are awarded rather than just whether you got the answer right.

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Radians Questions

  1. A Level
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