HSC Science (Electronics) 12th Board ExamMaharashtra State Board
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# Solution for The Maximum Velocity of a Particle Performing Linear S.H.M. is 0.16 m/s. If Its Maximum Acceleration is 0.64 m/s^2, Calculate Its Period. - HSC Science (Electronics) 12th Board Exam - Physics

ConceptDifferential Equation of Linear S.H.M.

#### Question

The maximum velocity of a particle performing linear S.H.M. is 0.16 m/s. If its maximum acceleration is 0.64 m/s2, calculate its period.

#### Solution

vmax = 0.16 m/s, amax =0.64 m/s2

Period (T) = ?
vmax = Aω and amax = Aω2

a_max/v_max=omega=(2pi)/T

T=(2pixxv_max)/a_max

=(2xx3.14xx0.16)/0.64

= 1.571 s
Period of particle is 1.571 s

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Solution for question: The Maximum Velocity of a Particle Performing Linear S.H.M. is 0.16 m/s. If Its Maximum Acceleration is 0.64 m/s^2, Calculate Its Period. concept: Differential Equation of Linear S.H.M.. For the courses HSC Science (Electronics), HSC Science (Computer Science), HSC Science (General)
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