AP EAMCET · PHYSICS · Laws of Motion
A 3 kg block is connected as shown in the figure. Spring constants of two springs \(\mathrm{K}_1\) and \(\mathrm{K}_2\) are \(50 \mathrm{Nm}^{-1}\) and \(150 \mathrm{Nm}^{-1}\) respectively. The block is released from rest with the springs unstretched, The acceleration of the block in its lowest position is \(\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)\)
- A \(10 \mathrm{~ms}^{-2}\)
- B \(12 \mathrm{~ms}^{-2}\)
- C \(8 \mathrm{~ms}^{-2}\)
- D \(8.8 \mathrm{~ms}^{-2}\)
Answer & Solution
Correct Answer
(A) \(10 \mathrm{~ms}^{-2}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{k}_1=50 \mathrm{Nm}^{-1}, \mathrm{k}_2=150 \mathrm{Nm}^{-1}\) \(\mathrm{W}=\sqrt{\frac{\mathrm{k}_{\mathrm{eq}}}{\mathrm{m}}}=\sqrt{\frac{\mathrm{k}_1+\mathrm{k}_2}{\mathrm{~m}}} \sqrt{\frac{50+150}{3}}=\sqrt{\frac{200}{3}} \mathrm{rad} / \mathrm{s}\) Amplitude,…
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