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Question
The motion of a particle executing simple harmonic motion is described by the displacement function,
x (t) = A cos (ωt + φ).
If the initial (t = 0) position of the particle is 1 cm and its initial velocity is ω cm/s, what are its amplitude and initial phase angle? The angular frequency of the particle is π s–1. If instead of the cosine function, we choose the sine function to describe the SHM: x = B sin (ωt + α), what are the amplitude and initial phase of the particle with the above initial conditions.
Solution 1
Initially, at t = 0:
Displacement, x = 1 cm
Initial velocity, v = ω cm/sec.
Angular frequency, ω = π rad/s–1
It is given that:
`x(t) = A cos(omegat + phi)`
`1 = A cos(omega xx 0 xx phi)`
`A cos phi = 1` ....(i)
`"Velocity", v = (dx)/(dt)`
`omega = - Aomegasin(omegat + phi)`
`1 = -A sin (omega xx 0 + phi)` = `- A Sin phi`
`Asin phi = -1` ...(ii)
Squaring and adding equations (i) and (ii), we get:
`A^2(sin^2 phi + cos^2 phi) = 1+ 1`
`A^2= 2`
`:. A = sqrt2 cm`
Dividing equation (ii) by equation (i), we get:
`tan phi = -1`
`: phi = (3pi)/4, (7pi)/4,....`
SHM is given as:
`x = B sin(omegat + alpha)`
Putting the given values in this equation, we get:
`1 = B sin[omegaxx0 + alpha]`
`B sin alpha = 1` ....(iii)
Velocity, `v = omegaBcos(omegat + alpha)`
Substituting the given values, we get:
`pi = piBsin alpha`
`Bsin alpha = 1` .....(iv)
Squaring and adding equations (iii) and (iv), we get:
`B^2[sin^2alpha + cos^2 alpha] = 1+ 1`
`B^2 = 2`
`:. B = sqrt2 cm`
Dividing equation (iii) by equation (iv), we get:
`(Bsin alpha)/(B cos alpha) = 1/1`
`tan alpha = 1 = tan pi/4`
`:. alpha = pi/4 , (5pi)/4, ......`
Solution 2
The given displacement function is
`x(t) = A cos (omegat + phi)` ....(i)
At t = 0, x(0) = 1 cm,Also `omega = pis^(-1)`
`:. 1 = A cos (pi xx 0 +phi)`
`=> A cos phi = 1`
Also, differentiating equation (i) w.r.t 't'
`v= d/(dt) x(t ) = -A omega sin (omegat + phi)`
Now at t = 0, `v = omega`
:. From equation (iii) `omega = -A omega sin (pi xx0+ phi)` or `A sinphi = -1`
Squaring and adding eqations (ii) and (iv)
`A^2 cos^2phi + A^2 sin^2 phi = 1^2 + 1^2` or `A = sqrt2 cm`
Dividing equation (ii) and (iv)
`(A sin phi)/(A cos phi) = (-1)/1`
`:. tan phi = -1 => phi = (3pi)/4`
If instead we use the sine function, i.e
`x = Bsin (omegat + alpha)` ,then `v = d/(dt) B omega cos (omegat + alpha)`
:. At t = 0 using x= 1 and `v = omega`, we get `1 = B sin(omega xx 0 + alpha)`
or `B sin alpha = 1` ....(v)
and `omega = Bomega cos (omega xx 0 + alpha)` or `B cos alpha = 1` ...(vi)
Dividing v and vi
`tan alpha = 1 or alpha = pi/4 or (5pi)/4`
Squaring v and vi, we get
`B^2 sin^2 alpha + B^2 cos^2 alpha = 1^2 + 1^2`
`=> B = sqrt2 cm`
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