MHT CET · Physics · Oscillations
A pendulum has length of \(0.4 \mathrm{~m}\) and maximum speed \(4 \mathrm{~m} / \mathrm{s}\). When the length makes
an angle \(30^{\circ}\) with the horizontal, its speed will be
\(\left[\sin \frac{\pi}{6}=\cos \frac{\pi}{3}=0 \cdot 5\right.\) and \(\left.\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right]\)
- A \(2 \sqrt{2} \mathrm{~m} / \mathrm{s}\)
- B \(\sqrt{3} \mathrm{~m} / \mathrm{s}\)
- C \(2 \sqrt{5} \mathrm{~m} / \mathrm{s}\)
- D \(2 \sqrt{3} \mathrm{~m} / \mathrm{s}\)
Answer & Solution
Correct Answer
(D) \(2 \sqrt{3} \mathrm{~m} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
A pendulum has length of \(0.4 \mathrm{~m}\) and maximum speed \(4 \mathrm{~m} / \mathrm{s}\). When the length makes an angle \(30^{\circ}\) with the horizontal, its speed will be \(2 \sqrt{3} \mathrm{~m} / \mathrm{s}\).
Explanation:
\(h=L-L \cos 0=L(1-\cos \theta)\)
\(=0.4\left(1-\cos 60^{\circ}\right)\)
\(=0.4\left(1-\frac{1}{2}\right)\)
\(=0.4 \times \frac{1}{2}\)
\(=0.2\)
\(\frac{1}{2} m v_{1}^{2}=\frac{1}{2} m v_{2}^{2}+m g h\)
\(v_{1}^{2}=v_{2}^{2}+2 g h\)
\(\therefore v_{2}^{2}=v_{1}^{2}-2 g h\)
\(=(4)^{2}-2 \times 10 \times 0.2\)
\(=16-4=12\)
\(\therefore v_{2}=\sqrt{12}=2 \sqrt{3} m / s\)
Explanation:
\(h=L-L \cos 0=L(1-\cos \theta)\)
\(=0.4\left(1-\cos 60^{\circ}\right)\)
\(=0.4\left(1-\frac{1}{2}\right)\)
\(=0.4 \times \frac{1}{2}\)
\(=0.2\)
\(\frac{1}{2} m v_{1}^{2}=\frac{1}{2} m v_{2}^{2}+m g h\)
\(v_{1}^{2}=v_{2}^{2}+2 g h\)
\(\therefore v_{2}^{2}=v_{1}^{2}-2 g h\)
\(=(4)^{2}-2 \times 10 \times 0.2\)
\(=16-4=12\)
\(\therefore v_{2}=\sqrt{12}=2 \sqrt{3} m / s\)
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