The motion of a particle executing simple harmonic motion is described by the displacement function, x(t) = A cos (ωt + ∅ ). 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 p 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.
Given: The equation of motion of a particle executes SHM is x ( t ) = A cos ( ω t + ϕ ) . At t = 0 the position of the particle is 1 cm . The initial velocity of the particle is ω cm / s and angular frequency of the particle is π s -1 .
The equation of SHM is given by,
x ( t ) = A cos ( ω t + ϕ ) (1)
Where, the displacement of particle is x ( t ) , the amplitude is A , the angular frequency is ω and the phase difference is ϕ .
According to given condition,
At t = 0 , x ( 0 ) = 1 cm
By substituting the value of t as zero in equation (1), we get
x ( 0 ) = A cos ( ω × 0 + ϕ ) 1 = A cos ϕ (2)
Differentiating equation (1) with respect to time, we get
By substituting the value of x ( t ) in equation (3), we get
d x ( t ) d t = d d t ( A cos ( ω t + ϕ ) ) v ( t ) = − A ω sin ( ω t + ϕ ) (3)
According to given condition,
At t = 0 , v ( 0 ) = ω cm/s
By substituting the given values in equation (3), we get
ω = − A ω sin ( ω × 0 + ϕ ) − 1 = A sin ϕ (4)
Square and add the equation (2) and equation (4).
A 2 ( cos 2 ϕ + sin 2 ϕ ) = 1 + 1 A 2 = 2 A = 2 cm
Divide equation (4) by equation (2).
tan ϕ = − 1 tan ϕ = tan ( 3 π 4 ) , tan ( 7 π 4 ) , …
By comparing the terms on both the sides, we get
ϕ = 3 π 4 , 7 π 4 , …
Thus, the amplitude and initial phase angle are 2 cm and 3 π 4 , 7 π 4 , … respectively.
The general equation of Simple Harmonic Motion (SHM) is given by,
y = B sin ( ω t + α ) (5)
Substitute y = 1 at x = 0 in equation (5).
1 = B sin ( ω × 0 + α ) 1 = B sin α (6)
Differentiate the equation (5) with respect to time.
v = d d t ( B sin ( ω t + α ) ) = B ω cos ( ω t + α )
Substitute v = ω at t = 0 in above equation.
ω = B ω cos ( ω × 0 + α ) 1 = B cos α (7)
Square and add the equation (6) and equation (7).
B 2 ( cos 2 ϕ + sin 2 ϕ ) = 1 + 1 B 2 = 2 B = 2 cm
Divide equation (6) by equation (7).
B sin α B cos α = 1 tan α = 1 tan α = tan ( π 4 , 5 π 4 , … ) α = π 4 , 5 π 4 , …
Thus, the amplitude and initial phase angle are 2 cm and π 4 , 5 π 4 , … respectively.
Given: The equation of motion of a particle executes SHM is
The equation of SHM is given by,
Where, the displacement of particle is
According to given condition,
At
By substituting the value of
Differentiating equation (1) with respect to time, we get
By substituting the value of
According to given condition,
At
By substituting the given values in equation (3), we get
Square and add the equation (2) and equation (4).
Divide equation (4) by equation (2).
By comparing the terms on both the sides, we get
Thus, the amplitude and initial phase angle are
The general equation of Simple Harmonic Motion (SHM) is given by,
Substitute
Differentiate the equation (5) with respect to time.
Substitute
Square and add the equation (6) and equation (7).
Divide equation (6) by equation (7).
Thus, the amplitude and initial phase angle are
