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.
Answers
Explanation:
Intially, at t = 0;
Displacement, x = 1 cm
Intial velocity, v = ω cm/ sec.
Angular frequency, ω = π rad/s–1
It is given that,
x(t) = A cos(ωt + Φ)
1 = A cos(ω × 0 + Φ) = A cos Φ
A cosΦ = 1 ...(i)
Velocity, v= dx/dt
ω = -A ωsin(ωt + Φ)
1 = -A sin(ω × 0 + Φ) = -A sin Φ
A sin Φ = -1 ...(ii)
Squaring and adding equations (i) and (ii), we get:
A2 (sin2 Φ + cos2 Φ) = 1 + 1
A2 = 2
∴ A = √2 cm
Dividing equation (ii) by equation (i), we get:
tanΦ = -1
∴ Φ = 3π/4 , 7π/4,......
SHM is given as:
x = Bsin (ωt + α)
Putting the given values in this equation, we get:
1 = Bsin[ω × 0 + α] = 1 + 1
Bsin α = 1 ...(iii)
Velocity, v = ωBcos (ωt + α)
Substituting the given values, we get:
π = πBsin α
Bsin α = 1 ...(iv)
Squaring and adding equations (iii) and (iv), we get:
B2 [sin2 α + cos2 α] = 1 + 1
B2 = 2
∴ B = √2 cm
Dividing equation (iii) by equation (iv), we get:
Bsin α / Bcos α = 1/1
tan α = 1 = tan π/4
∴ α = π/4, 5π/4,......
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Given,
X = Acos(wt + ø)
at t = 0 position of particle is 1 cm and velocity = w cm/s
so , 1 = Acos( w×0 + ø)
1 = Acosø ------(1)
again,
x = Acos(wt + ø)
differentiate wrt t
dx/dt = -Awsin(wt + ø)
at t = 0 , V = w cm/s
So, w = -Awsin(w×0 +ø)
1 = -Asinø ------(2)
From eqns (1) and(2)
2 = A²( cos²ø + sin²ø)
A = ±√2
Hence, amplitude {A} = √2 cm
Again,
Divide eqns (2) ÷ (1)
tanø = -1
ø = -π/4 or 7π/4
Now ,if we choose
X = Bsin(wt + a )
Then, at t = 0 ,
X = 1 cm
V = w cm/s
Put t = 1 and x = 1 in above equation .
1 = Bsina ------(1)
and
dx/dt = Bwcos(wt +a)
Put t = 0 and V = w
1 = Bcosa -----(2)
From eqns (1) and (2)
B = ±√2
Hence amplitude = √2 cm
Tana = 1
a = π/4