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.
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Answer 1

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,......

Answer 2

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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

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