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Sagot :
Let's solve the problem step-by-step:
### Part (i)
Given:
- [tex]\( y_1 = 25 \sin (15t - 6x) \)[/tex]
- [tex]\( y_2 = 25 \sin (15t + 6x) \)[/tex]
We need to find the resulting wave [tex]\(y_{\text{result}} = y_1 + y_2\)[/tex].
To do this, we use a trigonometric identity:
[tex]\[ \sin(A) + \sin(B) = 2 \sin\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) \][/tex]
For our waves:
- Let [tex]\( A = 15t - 6x \)[/tex]
- Let [tex]\( B = 15t + 6x \)[/tex]
Plugging these into the identity, we get:
[tex]\[ y_1 + y_2 = 25 \sin (15t - 6x) + 25 \sin (15t + 6x) \][/tex]
Let's express it using the trigonometric identity:
[tex]\[ \sin(15t - 6x) + \sin(15t + 6x) = 2 \sin\left( \frac{(15t - 6x) + (15t + 6x)}{2} \right) \cos\left( \frac{(15t - 6x) - (15t + 6x)}{2} \right) \][/tex]
[tex]\[ = 2 \sin\left( \frac{30t}{2} \right) \cos\left( \frac{-12x}{2} \right) \][/tex]
[tex]\[ = 2 \sin(15t) \cos(6x) \][/tex]
Thus,
[tex]\[ 25 \sin(15t - 6x) + 25 \sin(15t + 6x) = 2 \times 25 \sin(15t) \cos(6x) \][/tex]
[tex]\[ = 50 \sin(15t) \cos(6x) \][/tex]
So, the resulting wave is:
[tex]\[ y_{\text{result}} = 50 \sin (15t) \cos (6x) \][/tex]
### Part (ii)
Given:
- [tex]\( y_1 = 4 \cos (2t - 30) \)[/tex]
- [tex]\( y_2 = -3 \sin (2t + 60) \)[/tex]
We need to find the resulting wave [tex]\(y_{\text{result}} = y_1 + y_2\)[/tex].
So the resulting wave is simply the sum of the two given displacements:
[tex]\[ y_{\text{result}} = 4 \cos (2t - 30) + (-3 \sin (2t + 60)) \][/tex]
[tex]\[ y_{\text{result}} = 4 \cos (2t - 30) - 3 \sin (2t + 60) \][/tex]
In conclusion, the resulting waves are:
1. For part (i):
[tex]\[ y_{\text{result}} = 50 \sin (15t) \cos (6x) \][/tex]
2. For part (ii):
[tex]\[ y_{\text{result}} = 4 \cos (2t - 30) - 3 \sin (2t + 60) \][/tex]
### Part (i)
Given:
- [tex]\( y_1 = 25 \sin (15t - 6x) \)[/tex]
- [tex]\( y_2 = 25 \sin (15t + 6x) \)[/tex]
We need to find the resulting wave [tex]\(y_{\text{result}} = y_1 + y_2\)[/tex].
To do this, we use a trigonometric identity:
[tex]\[ \sin(A) + \sin(B) = 2 \sin\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) \][/tex]
For our waves:
- Let [tex]\( A = 15t - 6x \)[/tex]
- Let [tex]\( B = 15t + 6x \)[/tex]
Plugging these into the identity, we get:
[tex]\[ y_1 + y_2 = 25 \sin (15t - 6x) + 25 \sin (15t + 6x) \][/tex]
Let's express it using the trigonometric identity:
[tex]\[ \sin(15t - 6x) + \sin(15t + 6x) = 2 \sin\left( \frac{(15t - 6x) + (15t + 6x)}{2} \right) \cos\left( \frac{(15t - 6x) - (15t + 6x)}{2} \right) \][/tex]
[tex]\[ = 2 \sin\left( \frac{30t}{2} \right) \cos\left( \frac{-12x}{2} \right) \][/tex]
[tex]\[ = 2 \sin(15t) \cos(6x) \][/tex]
Thus,
[tex]\[ 25 \sin(15t - 6x) + 25 \sin(15t + 6x) = 2 \times 25 \sin(15t) \cos(6x) \][/tex]
[tex]\[ = 50 \sin(15t) \cos(6x) \][/tex]
So, the resulting wave is:
[tex]\[ y_{\text{result}} = 50 \sin (15t) \cos (6x) \][/tex]
### Part (ii)
Given:
- [tex]\( y_1 = 4 \cos (2t - 30) \)[/tex]
- [tex]\( y_2 = -3 \sin (2t + 60) \)[/tex]
We need to find the resulting wave [tex]\(y_{\text{result}} = y_1 + y_2\)[/tex].
So the resulting wave is simply the sum of the two given displacements:
[tex]\[ y_{\text{result}} = 4 \cos (2t - 30) + (-3 \sin (2t + 60)) \][/tex]
[tex]\[ y_{\text{result}} = 4 \cos (2t - 30) - 3 \sin (2t + 60) \][/tex]
In conclusion, the resulting waves are:
1. For part (i):
[tex]\[ y_{\text{result}} = 50 \sin (15t) \cos (6x) \][/tex]
2. For part (ii):
[tex]\[ y_{\text{result}} = 4 \cos (2t - 30) - 3 \sin (2t + 60) \][/tex]
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