(Sine)ing

Calculus Level 2

A particle is executing 1D motion. Its acceleration as a function of time is given by a = − A w 2 sin ⁡ ( w t ) a=-Aw^{2}\sin(wt) . At time t t , the magnitude of displacement from the mean position, velocity, acceleration are equal. Then find the value of t t .

Details and assumptions:

  • At maximum acceleration(magnitude), velocity is 0
  • At maximum velocity, displacement is zero
  • w w is angular frequency
  • 'A' a is constant of appropriate dimensions
3 3 π 2 \frac{\pi}{2} π \pi π 4 \frac{\pi}{4}

Sparsh Sarode by Sparsh Sarode , 22, India

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1 solution

Sparsh Sarode
May 17, 2016

d v d t = − A w 2 s i n ( w t ) \frac{dv}{dt}=-Aw^{2}sin(wt)

v = A w c o s ( w t ) v=Awcos(wt)

x = A s i n ( w t ) x=Asin(wt)

Since magnitude of acceleration, displacement, velocity are equal,

A w 2 s i n ( w t ) = A s i n ( w t ) ⇒ w = 1 Aw^{2}sin(wt)=Asin(wt) \Rightarrow w=1

Equating velocity and displacement and Substituting w = 1 w=1 ,

s i n t = c o s t ⇒ t = π 4 sint=cost \Rightarrow t=\frac{π}{4}

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