Question

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 $πs_{−1}$. If instead of the cosine function we choose the sine function to describe the $SHM:x=Bsin(ωt+α)$, what are the amplitude and initial phase of the particle with the above initial conditions.

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Updated on : 2022-09-05

Solution

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Displacement, $x=1cm$

Intial velocity, $v=ωcm/sec.$

Angular frequency, $ω=πrad/s_{−1}$

It is given that,

$x(t)=Acos(ωt+ϕ)$

$1=Acos(ω×0+ϕ)=Acosϕ$

$Acosϕ=1$ ...(i)

Velocity, $v=dx/dt$

$ω=−Aωsin(ωt+ϕ)$

$1=−Asin(ω×0+ϕ)=−Asinϕ$

$Asinϕ=−1$ ...(ii)

Squaring and adding equations (i) and (ii), we get:

$A_{2}(sin_{2}ϕ+cos_{2}ϕ)=1+1$

$A_{2}=2$

$A_{2}=2$

$∴A=2 cm$

Dividing equation (ii) by equation (i), we get:

Dividing equation (ii) by equation (i), we get:

$tanϕ=−1$

$∴ϕ=3π/4,7π/4$

SHM is given as:

$∴ϕ=3π/4,7π/4$

SHM is given as:

$x=Bsin(ωt+α)$

Putting the given values in this equation, we get:

Putting the given values in this equation, we get:

$1=Bsin(ω×0+α]=1+1$

$Bsinα=1$ ...(iii)

Velocity, $v=ωcos(ωt+α)$

Substituting the given values, we get:

$Bsinα=1$ ...(iii)

Velocity, $v=ωcos(ωt+α)$

Substituting the given values, we get:

$π=πBcosα$

$Bcosα=1$ ...(iv)

Squaring and adding equations(iii) and (iv), we get:

Squaring and adding equations(iii) and (iv), we get:

$B_{2}[sin_{2}α+cos_{2}α]=1+1$

$B_{2}=2$

$B_{2}=2$

$∴B=2 $

Dividing equation (iii) by equation (iv), we get:

Dividing equation (iii) by equation (iv), we get:

$Bsinα/Bcosα=1/1$

$tanα=1=tanπ/4$

$∴α=π/4,5π/4,$......

$tanα=1=tanπ/4$

$∴α=π/4,5π/4,$......

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