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Mechanical Properties Of Fluids

Question
CBSEENPH11020200

The motion of a particle executing simple harmonic motion is described by the displacement function,
 
x (t) = cos (ω+ φ).  

If the initial (= 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.

Solution
Intially, at t = 0; 
Displacement, x = 1 cm 
Initial 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, vdx/dt 
ω = -A ωsin(ωt + Φ
1 = -A sin(ω × 0 + Φ) = -sin Φ 

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,......