Curly waves

Classical Mechanics Level 3

Equation of two waves are
y = A sin ⁡ ( w t − k x ) y = A \sin(wt - kx)
y = A sin ⁡ ( w t − k x + 90 ) y=A \sin(wt - kx + 90) 90 denotes 90 degrees,
A is amplitude,w is angular frequency, k k is wave number;

Now the equation of the resulting wave can be represented as
y = A a 2 sin ⁡ ( w t − k x + b ) y = A\sqrt [ 2 ]{ a } \sin(wt - kx + b) where a a is an integer, and b b is in degrees
Find ( b ∗ 2 ) / ( a ∗ 5 ) (b*2)/(a*5) .

This is a part of my set Aniket's Mechanics Challenges


The answer is 9.

Aniket Sanghi by Aniket Sanghi , 21, India

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1 solution

Neelesh Vij
Mar 14, 2016

The resultant wave equation can be written as:

y = A sin ⁡ ( w t − k x ) + A sin ⁡ ( w t − k x + 9 0 ∘ ) y= A\sin{(wt -kx)} + A\sin{(wt -kx +90^{\circ})}

Using identity sin ⁡ A + sin ⁡ B = 2 sin ⁡ ( A + B 2 ) × cos ⁡ ( A − B 2 ) \text{Using identity} \sin{A} + \sin{B} = 2\sin{\left(\dfrac{A+B}{2} \right)} \times \cos{\left(\dfrac{A-B}{2} \right)}

→ 2 A sin ⁡ ( ( k x − w t ) + ( k x − w t + 9 0 ∘ ) 2 ) × cos ⁡ ( ( k x − w t ) − ( k x − w t + 9 0 ∘ ) 2 ) \displaystyle \rightarrow 2A\sin{\left(\dfrac{(kx -wt)+ (kx -wt +90^{\circ})}{2} \right)} \times \cos{\left(\dfrac{(kx -wt)-(kx -wt +90^{\circ})}{2} \right)}

→ 2 A sin ⁡ ( w t − k x + 4 5 ∘ ) × cos ⁡ 4 5 ∘ \rightarrow 2A \sin{(wt-kx + 45^{\circ})} \times \cos{45^{\circ}}

→ A 2 × sin ⁡ ( w t − k x + 4 5 ∘ ) \rightarrow A\sqrt{2} \times \sin{(wt-kx +45^{\circ})}

∴ a = 2 and b = 45 \therefore a =2 \text{and} b=45

→ b × 2 a × 5 = 45 × 2 2 × 5 = 9 \rightarrow \dfrac{b\times 2}{a \times5} = \dfrac{45 \times 2}{2\times 5} = \boxed{9}

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