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A particle executes SHM with amplitude A = 5 cm and time period T = 2 s — JEE Physics

SDSiddharth Das · 11 Asked 2mo ago 246 views 1 answer

A particle executes SHM with amplitude $A = 5$ cm and time period $T = 2$ s. Find the maximum velocity and maximum acceleration.

1 Answer

VAVidaara Admin ✓ Vidaara Team ✓ Accepted · 2mo ago ▲ 39

$$\omega = \frac{2\pi}{T} = \pi rad/s$$

$$v_{max} = A\omega = 5\pi cm/s \approx 15.7 cm/s$$

$$a_{max} = A\omega^2 = 5\pi^2 cm/s^2 \approx 49.3 cm/s^2$$

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Discussion (4)

GP
Why do we take the positive value only in the last step?
Gaurav Pandey · 2mo ago
SJ
Why do we take the positive value only in the last step?
Shruti Jain · 1mo ago
AK
Saved me before my mock test. Much clearer than my coaching notes.
Aditya Kumar · 1mo ago
MT
Can someone explain why we ignore the other root here?
Manish Tiwari · 1mo ago
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