Question

Two SHM's are
X1 (mm, sec) = 1.0 sin[π/4(12t+1)]
, and x2 (mm, sec) = 0.5 [sin 3πt + V3 cos 3πt]
Discuss the ratio of their amplitudes, frequency and find the phase difference.​

Answer 1

Solution :

y1=10sin(3πt+π/4)….(i)  

y2=5sin3π+t+53–√cost ….(ii)  

In Eq.(ii), let  

5=acosθ…(iii)  

and 53–√=asinθ….(iv)  

∴y2=a sin 3πt cos2θ + asinθcos3πt  

y2=asin(3πt+θ)....(v)  

Squaring and adding Eqs. (iii) and (iv), we get  

52+(53–√)2=a2cos2θ+a2sin2θ  

⇒100=a2  

⇒a=10  

Therefore, Eqs.(v) cab be written as  

y2=10sin(3πt+θ)....(vi)  

From Eqs (i) and (vi), the ratio of amplitudes is 10:10, i.e., 1:1.

Answer 2

Answer:

X1 mm sec is 1.0

X2 mm sec is 0.5

ratio is 1.0/0.5

1:5