Simple Harmonic question

Question

A particle in shm has a time period of 4s. It takes time t1 to start from mean position and reach half of the amplitude.
In another case, it takes a time t2 to start from extreme position and reach half the amplitude ,then options

1. t1/t2=1
2. t1/t2=1/2

Answer 1

Hi Prashant!

 

Simple harmonic motion can be described by a time-varying sinusoid (meaning either a sine wave or a cosine wave; they are the same shape, just shifted with respect to each other).

 

With the proper setup in the argument of the function, any SHM can be described by either a sine or cosine function.  However, it is usually most straightforward to choose the one which corresponds to the initial conditions you have (at t = 0).

 

For case 1, where the object starts at the mean position, we need position (x) = 0 at t = 0.  This means we want to use a sine wave, of the form:

 

x(t) = Asin((2π/T)t)

 

where T is the period of the SHM (though this turns out not to matter in the solution) and A is the amplitude.  This function has the attribute that x = 0 when t = 0, signifying that the object starts at "the mean position."

 

For the position x to be half of the amplitude at time t₁, we need:

 

x(t₁) = (A/2) = Asin((2π/T)t₁)

 

This requires that sin((2π/T)t₁) = 1/2.

 

Starting from 0 an going toward positive angles, sin(θ) = 1/2 first when θ = 30^o = π/6 rad.  So, for sin((2π/T)t₁) = 1/2, we need:

 

(2π/T)t₁ = π/6

 

We can make the same procedure for starting from the extreme position, except, in this case, it is easier to represent x(t) with a cosine function, since cos(0) = 1.  This means if we define:

 

x(t) = Acos((2π/T)t)

 

then x(0) = A, signifying that the object starts at the extreme position (at t = 0).  

 

Making the same argument as above, we are looking for the first value of the argument of the cosine function that makes cos((2π/T)t₂) = 1/2.  Starting from 0 and increasing, cos(θ) = 0 first when θ = 60^o = π/3 rad.  So we need:

 

(2π/T)t₂ = π/3

 

Now we can make the comparison you need:

 

(2π/T)t₁       (π/6)

----------- = --------

(2π/T)t₂        (π/3)

 

[This shows, by the way, why the value of the period does not matter.  It will cancel out, no matter what it is.  This calculation will thus be true for ANY simple harmonic motion]

 

This leads, in a step, to one of your options above.

 

The answer makes sense with the physical aspects of a system that follows Hooke's law (and thus executes simple harmonic motion).  When the object is out at its extreme position, it experiences a greater force, and thus has a greater acceleration, meaning it speeds up and covers ground more quickly out there.  So it should take less time to go from the extreme position to half-amplitude, than to go from the mean position to half-amplitude.

 

I hope this helps get you on your way!  Look this over, and let me know if you want to talk about any aspects of it further.