Question

A particle performing linear SHM at amplitude of 8cm and period at 45. find the time taken by the describe a distance at 2cm from the positive extreme position​

Answer 1

Answer:

given, period , T = 12 sec

Amplitude , A = 8cm

so, angular frequency,\omegaω = 2π/T = 2π/12 = π/6 rad/s

When the particle covers a distance of 6cm from the positive extremity. Its displacement from the mean position is x = 8-6 = 2cm

We know, equation of SHM is given by

x=Acos\omega tx=Acosωt [from extreme position, ]

x =8sin(π/6)t

at x = 2cm

2 = 8cos(π/6)t

1/4 = cos(π/6)t

75°52' × π/180° = (π/6) × t

t = 2.517 sec

Explanation:

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Answer 2

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