355083
Equation of motion in the same direction is given by \(y_{1}=A \sin (\omega t-k x), \quad y_{2}=A \sin (\omega t-k x-\theta)\). The amplitudeof the medium particle will be
1 \(\sqrt{2} A \cos \theta\)
2 \(2 A \cos \theta\)
3 \(2 A \cos \dfrac{\theta}{2}\)
4 \(\sqrt{2} A \cos \dfrac{\theta}{2}\)
Explanation:
The resultant amplitude is given by \(\begin{gathered}A_{r}=\sqrt{A^{2}+A^{2}+2 A A \cos \theta}=\sqrt{2 A^{2}(1+\cos \theta)} \\A_{r}=2 A \cos \dfrac{\theta}{2} \quad\left(\because 1+\cos \theta=2 \cos ^{2} \dfrac{\theta}{2}\right)\end{gathered}\)
PHXI15:WAVES
355084
Two waves having intensities \(25: 9\) produce interference. The ratio maximum to minimum intensity is equal to
355085
Two periodic waves of amplitude \(A_{1}\) and \(A_{2}\) pass through a region. If \(A_{1}>A_{2}\) the difference in the maximum and minimum resultant amplitude possible is
355086
The equation of two waves are given by: \({y_1} = 5\sin 2\pi (x - vt)cm\) \({y_2} = 3\sin 2\pi (x - vt + 1.5)cm\) These waves are simultaneuously passing through a string. The amplitude of the resulting wave is
1 \(2\;cm\)
2 \(4\;cm\)
3 \(5.8\;cm\)
4 \(8\;cm\)
Explanation:
\({A_1} = 5\;cm,{A_2} = 3\;cm\) The two waves differ in their phases by \(\Delta \phi=2 \pi(1.5)=3 \pi\) radians Required resulting 'amplitude' is: \(A_{\text {net }}=\sqrt{A_{1}^{2}+A_{2}^{2}+2 A_{1} A_{2} \cos (\Delta \phi)}\) \(=\sqrt{5^{2}+3^{2}+2 \times 5 \times 3 \cos 3 \pi}\) \( = \sqrt {25 + 9 - 1} = \sqrt {33} \approx 5.8\;cm\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI15:WAVES
355083
Equation of motion in the same direction is given by \(y_{1}=A \sin (\omega t-k x), \quad y_{2}=A \sin (\omega t-k x-\theta)\). The amplitudeof the medium particle will be
1 \(\sqrt{2} A \cos \theta\)
2 \(2 A \cos \theta\)
3 \(2 A \cos \dfrac{\theta}{2}\)
4 \(\sqrt{2} A \cos \dfrac{\theta}{2}\)
Explanation:
The resultant amplitude is given by \(\begin{gathered}A_{r}=\sqrt{A^{2}+A^{2}+2 A A \cos \theta}=\sqrt{2 A^{2}(1+\cos \theta)} \\A_{r}=2 A \cos \dfrac{\theta}{2} \quad\left(\because 1+\cos \theta=2 \cos ^{2} \dfrac{\theta}{2}\right)\end{gathered}\)
PHXI15:WAVES
355084
Two waves having intensities \(25: 9\) produce interference. The ratio maximum to minimum intensity is equal to
355085
Two periodic waves of amplitude \(A_{1}\) and \(A_{2}\) pass through a region. If \(A_{1}>A_{2}\) the difference in the maximum and minimum resultant amplitude possible is
355086
The equation of two waves are given by: \({y_1} = 5\sin 2\pi (x - vt)cm\) \({y_2} = 3\sin 2\pi (x - vt + 1.5)cm\) These waves are simultaneuously passing through a string. The amplitude of the resulting wave is
1 \(2\;cm\)
2 \(4\;cm\)
3 \(5.8\;cm\)
4 \(8\;cm\)
Explanation:
\({A_1} = 5\;cm,{A_2} = 3\;cm\) The two waves differ in their phases by \(\Delta \phi=2 \pi(1.5)=3 \pi\) radians Required resulting 'amplitude' is: \(A_{\text {net }}=\sqrt{A_{1}^{2}+A_{2}^{2}+2 A_{1} A_{2} \cos (\Delta \phi)}\) \(=\sqrt{5^{2}+3^{2}+2 \times 5 \times 3 \cos 3 \pi}\) \( = \sqrt {25 + 9 - 1} = \sqrt {33} \approx 5.8\;cm\)
355083
Equation of motion in the same direction is given by \(y_{1}=A \sin (\omega t-k x), \quad y_{2}=A \sin (\omega t-k x-\theta)\). The amplitudeof the medium particle will be
1 \(\sqrt{2} A \cos \theta\)
2 \(2 A \cos \theta\)
3 \(2 A \cos \dfrac{\theta}{2}\)
4 \(\sqrt{2} A \cos \dfrac{\theta}{2}\)
Explanation:
The resultant amplitude is given by \(\begin{gathered}A_{r}=\sqrt{A^{2}+A^{2}+2 A A \cos \theta}=\sqrt{2 A^{2}(1+\cos \theta)} \\A_{r}=2 A \cos \dfrac{\theta}{2} \quad\left(\because 1+\cos \theta=2 \cos ^{2} \dfrac{\theta}{2}\right)\end{gathered}\)
PHXI15:WAVES
355084
Two waves having intensities \(25: 9\) produce interference. The ratio maximum to minimum intensity is equal to
355085
Two periodic waves of amplitude \(A_{1}\) and \(A_{2}\) pass through a region. If \(A_{1}>A_{2}\) the difference in the maximum and minimum resultant amplitude possible is
355086
The equation of two waves are given by: \({y_1} = 5\sin 2\pi (x - vt)cm\) \({y_2} = 3\sin 2\pi (x - vt + 1.5)cm\) These waves are simultaneuously passing through a string. The amplitude of the resulting wave is
1 \(2\;cm\)
2 \(4\;cm\)
3 \(5.8\;cm\)
4 \(8\;cm\)
Explanation:
\({A_1} = 5\;cm,{A_2} = 3\;cm\) The two waves differ in their phases by \(\Delta \phi=2 \pi(1.5)=3 \pi\) radians Required resulting 'amplitude' is: \(A_{\text {net }}=\sqrt{A_{1}^{2}+A_{2}^{2}+2 A_{1} A_{2} \cos (\Delta \phi)}\) \(=\sqrt{5^{2}+3^{2}+2 \times 5 \times 3 \cos 3 \pi}\) \( = \sqrt {25 + 9 - 1} = \sqrt {33} \approx 5.8\;cm\)
355083
Equation of motion in the same direction is given by \(y_{1}=A \sin (\omega t-k x), \quad y_{2}=A \sin (\omega t-k x-\theta)\). The amplitudeof the medium particle will be
1 \(\sqrt{2} A \cos \theta\)
2 \(2 A \cos \theta\)
3 \(2 A \cos \dfrac{\theta}{2}\)
4 \(\sqrt{2} A \cos \dfrac{\theta}{2}\)
Explanation:
The resultant amplitude is given by \(\begin{gathered}A_{r}=\sqrt{A^{2}+A^{2}+2 A A \cos \theta}=\sqrt{2 A^{2}(1+\cos \theta)} \\A_{r}=2 A \cos \dfrac{\theta}{2} \quad\left(\because 1+\cos \theta=2 \cos ^{2} \dfrac{\theta}{2}\right)\end{gathered}\)
PHXI15:WAVES
355084
Two waves having intensities \(25: 9\) produce interference. The ratio maximum to minimum intensity is equal to
355085
Two periodic waves of amplitude \(A_{1}\) and \(A_{2}\) pass through a region. If \(A_{1}>A_{2}\) the difference in the maximum and minimum resultant amplitude possible is
355086
The equation of two waves are given by: \({y_1} = 5\sin 2\pi (x - vt)cm\) \({y_2} = 3\sin 2\pi (x - vt + 1.5)cm\) These waves are simultaneuously passing through a string. The amplitude of the resulting wave is
1 \(2\;cm\)
2 \(4\;cm\)
3 \(5.8\;cm\)
4 \(8\;cm\)
Explanation:
\({A_1} = 5\;cm,{A_2} = 3\;cm\) The two waves differ in their phases by \(\Delta \phi=2 \pi(1.5)=3 \pi\) radians Required resulting 'amplitude' is: \(A_{\text {net }}=\sqrt{A_{1}^{2}+A_{2}^{2}+2 A_{1} A_{2} \cos (\Delta \phi)}\) \(=\sqrt{5^{2}+3^{2}+2 \times 5 \times 3 \cos 3 \pi}\) \( = \sqrt {25 + 9 - 1} = \sqrt {33} \approx 5.8\;cm\)
