A pendulum of length 1 𝑚 is releasedfrom 𝜃 = 60°. The rate of change ofspeed of the bob at 𝜃 = 30° is (𝑔 =10 𝑚𝑠−2).(a) 10 𝑚𝑠−2(b) 7.5 𝑚𝑠−2(c) 5 𝑚𝑠−2(d) 5√3𝑚𝑠−2

Question

(a) 10 $\text{ ms}^{-2}$
(b) 7.5 $\text{ ms}^{-2}$
(c) 5 $\text{ ms}^{-2}$
(d) 5 $\sqrt{3} \text{ ms}^{-2}$

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The rate of change of speed of the bob at 𝜃 = 30° can be calculated using the formula for the acceleration of a pendulum, which is a = g * cos(𝜃).

Step 1: Convert the angle 𝜃 from degrees to radians. In this case, 𝜃 = 30° = π/6 radians.

Step 2: Substitute the values into the formula. a = 10 m/s² * cos(π/6).

Step 3: Calculate the cosine of π/6, which is √3/2.

Step 4: Multiply the values to get the acceleration. a = 10 m/s² * √3/2 = 5√3 m/s².

So, the rate of change of speed of the bob at 𝜃 = 30° is (d) 5√3 m/s².

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