Question

the maximum displacement of an oscillating particle is 0.05 .If its time period is 1.57 .what is its acceleration at the extreme position?​

Answer 1

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Given Time period, T = 1.57s

we know that,

angular frequency, ω = 2π/T

=> ω = 2 x 3.14/1.57 = 4

The maximum displacement = a = 0.05 m

The velocity will be maximum at the mean position, Hence velocity at the mean position,

Vmax = aω = 0.05 x 4 = 0.2 m/s

Hence velocity at the mean position will be 0.2 m/s

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Comments (1) Report

Could you please say acceleration at extreme position

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NlKlTA

NlKlTA Ambitious

Answer =>

The velocity at the mean position will be 0.2 m/s

Explanation =>

Given Time period, T = 1.57s

we know that,

angular frequency, ω = 2π/T

=> ω = 2 x 3.14/1.57 = 4

The maximum displacement = a = 0.05 m

The velocity will be maximum at the mean

position, Hence velocity at the mean position,

Vmax = aω = 0.05 x 4 = 0.2 m/s

Hence velocity at the mean position will be 0.2 m/s

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