### Q.7:- The motion of a particle executing simple harmonic motion is described by the displacement function,

*x* (*t*) = *A *cos (ω*t *+ 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.

**Answer:-**__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(ω

*t*+

*Φ*)

*A*cos(ω × 0 +

*Φ*) =

*A*cos

*Φ*

*A*cos

*Φ*= 1 …

**(i)**

*v*=

*dx*/

*dt*

*A*ωsin(ω

*t*+

*Φ*)

*A*sin(ω × 0 +

*Φ*) = –

*A*sin

*Φ*

*A*sin

*Φ*= -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 cm

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

tan*Φ *= -1

∴ *Φ *= 3π/4 , 7π/4,……

SHM is given as:

*x* = *B*sin (ω*t* + α)

Putting the given values in this equation, we get:

1 = *B*sin[ω × 0 + α] = 1 + 1

*B*sin α = 1 …**(iii)**

Velocity, v = ω*B*cos (ω*t* + α)

Substituting the given values, we get:

π = π*B*sin α

*B*sin α = 1 …**(iv)**

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

*B*^{2} [sin^{2} α + cos^{2} α] = 1 + 1

*B*^{2} = 2

∴ *B* = √2 cm

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

*B*sin α / *B*cos α = 1/1

tan α = 1 = tan π/4

∴ α = π/4, 5π/4,……