One end of a spring of force constant k is fixed to a vertical wall and the other to a…
Physics · JEE Advanced · NTA Exams — Oscillations and Waves
One end of a spring of force constant k is fixed to a vertical wall and the other to a body of mass m resting on a smooth horizontal surface. There is another wall at a distance x0 from the body. The spring is then compressed by 3x0 and released. The time taken to strike the wall from the instant of release is
(given sin–1(1/3) = (π/9))
(given sin–1(1/3) = (π/9))
- \(\frac{\pi}{6} \sqrt{\frac{m}{k}}\)
- \(\frac{2 \pi}{3} \sqrt{\frac{m}{k}}\)
- \(\frac{\pi}{4} \sqrt{\frac{m}{k}}\)
- \(\frac{11 \pi}{18} \sqrt{\frac{m}{k}}\)
Answer
(D) 11 18 m k
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