Physics, asked by Dhika, 1 year ago

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

Answered by Aastha6878

$\huge \mathfrak \red{solution}$

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,......

Answered by Ahaan6417

Explanation: