Question

explain the answer. ​

Answer 1

Answer:

i).

$A = 7.935853753$

ii).

$\delta = 54.38510262 degree \\ \delta = \frac{\pi}{3.30973} rads$

Explanation:

A]

$Given: \\ x1 = 5 \sin(4\pit + \frac{\pi}{3} ) \\ x2 = 3 \sin(4\pit + \frac{\pi}{4} )$

When two waves are superimposed resultant Amplitude is given by:

$A = \sqrt{ {a1}^{2} + {a2}^{2} + 2(a1)(a2) \cos( \alpha 1 - \alpha 2) }$

Where;

$a1 and a2 are amplitude of x1 and x2 respectively. \\ \alpha 1 \: and \: \alpha 2 \: are phase angle of x1 and x2 respectively.$

Here,

$a1 = 5 \: and \: a2 = 3 \\ \alpha 1 = \frac{\pi}{3} \: and \: \alpha 2 = \frac{\pi}{4}$

Therefore,

$A = \sqrt{ {5}^{2} + {3}^{2} + 2(5)(3) \cos(( \frac{\pi}{3}) - ( \frac{\pi}{4}) ) } \\ A = 7.935853753$

B].

Now for superimposed waves resultant epoch is given as:

$\delta = {tan}^{ - 1} ( \frac{a1 \sin( \alpha 1) + a2 \sin( \alpha 2) }{a1 \cos( \alpha 1) + a2c \cos( \alpha 2) } )$

$\delta = {tan}^{ - 1} ( \frac{5 \sin( \frac{\pi}{3} ) + 3 \sin( \frac{\pi}{4} ) }{5 \cos( \frac{\pi}{3} ) + 3 \cos( \frac{\pi}{4} ) } )$

$\delta = {tan}^{ - 1} (1.39608212) \\ \delta = 54.38510262 \: degree \\ \delta = \frac{\pi}{3.309729895} rads$