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My book says that,

For a simple harmonic motion, velocity of the particle
= -AωSin(ωt+φ)

Now, assuming φ to be π/2, in the next sentence, they say that,

maximum velocity = Aω.

Why have they simply taken the magnitude of velocity in this case?

Perhaps, they are talking about the magnitude of velocity, ie. speed.

But, what is the relevance of this negative sign ? Since velocity is a vector quantity, doesn't this equation imply that velocity is maximum when φ =(3π)/2 ? Does it have any significance?

Also, they use this equation without negative sign a number of times. Why? ie. can we use the equations -AωSin(ωt+φ) and AωSin(ωt+φ) interchangably? (Ofcourse not, but what factors decide which one to use?)

Say at this instant, the position of the particle is 3 m right of the mean position. Since, this particle is moving to the right, it should have a positive velocity (convention). Assuming φ to be zero, this equation -AωSin(ωt+φ) becomes -k, where k=AωSin(ωt).
So does it not mean that velocity is negative which contradics my earlier statement?

Please tell me where I have gone wrong. Thanks

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For a given oscillator, you can use either function to describe the component of the velocity in your chosen coordinate system, but the phase angle (which depends on when you start the clock) will be different. The magnitude of the maximum velocity will be Aω.

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  • $\begingroup$ @ R.W. Bird. Then why do we put a negative sign infront of such an equation if we could do calculation without it? $\endgroup$
    – Proxima
    Commented May 24, 2021 at 17:34
  • $\begingroup$ For instance, consider this: a body undergoing SHM follows the following equation :5cos(2πt+π/4). From this, one can write φ as π/4. But if I try to find its velocity, should I use - AωSin(ωt+φ) or AωSin(ωt+φ) with φ as π/4? Why? $\endgroup$
    – Proxima
    Commented May 24, 2021 at 17:40
  • $\begingroup$ To find the velocity at a particular time, put that time into the assumed function which has numbers. (maximum v = 5 and angles are in radians) $\endgroup$
    – R.W. Bird
    Commented May 24, 2021 at 19:20
  • $\begingroup$ R. W. Bird As you can see, ω= 2π. So at time=0, if I use the equation -AωSin(ωt+φ), i will get v= - 5*2πSin(π/4) which is a negative value. But if I use the equation AωSin(ωt+φ), I will get v=5*2πSin(π/4), which is a positive value. So, one of these is correct. Which one is correct and how am I supposed to know which one is that ? Is there any way? $\endgroup$
    – Proxima
    Commented May 25, 2021 at 7:51
  • $\begingroup$ Your original equation had no negative sign. $\endgroup$
    – R.W. Bird
    Commented May 25, 2021 at 20:36

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