A 100 g block is connected to a horizontal massless spring of force constant 25.6 N/m. As…
Physics · JEE Advanced · NTA Exams — Oscillations and Waves
A 100 g block is connected to a horizontal massless spring of force constant 25.6 N/m. As shown in figure (a ), the block is free to oscillate on a horizontal frictionless surface. The block is displaced 3 cm from the equilibrium position and, at t = 0, it is released from rest at x= 0 . It executes simple harmonic motion with the positive x-direction indicated in figure (a ).
The position-time(x-t) graph of motion of the block is as shown in figure (b ).
\[\text { (a) }\]
Position of the block as a function of time can now be expressed as
- \(x=3 \cos \left(16 t+\frac{\pi}{2}\right) \mathrm{cm}\)
- \(x=3 \cos \left(16 t+\frac{\pi}{3}\right) \mathrm{cm}\)
- \(x=3.5 \cos \left(16 t+\frac{\pi}{6}\right) \mathrm{cm}\)
- \(x=3.2 \cos \left(16 t+\frac{\pi}{4}\right) \mathrm{cm}\)
Answer
(C) x=3.5 (16 t+ 6 ) cm
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