355092
Two waves represented by \(y=a \sin (\omega t-k x)\) and \(y_{2}=\sqrt{3} a \cos (\omega t-k x)\) are superposed. The amplitude of the resultant wave is
1 \(2 a\)
2 \((2+\sqrt{3}) a\)
3 \((2-\sqrt{3}) a\)
4 \(2 \sqrt{3} a\)
Explanation:
Path difference \(=\pi / 2\) \(\therefore \quad a_{R}=\sqrt{a^{2}+(\sqrt{3} a)^{2}+2 \cdot a \cdot \sqrt{3} a \cdot \cos \pi / 2}=2 a\).
PHXI15:WAVES
355093
Two waves having equations \(x_{1}=a \sin \left(\omega t+\varphi_{1}\right), x_{2}=a \sin \left(\omega t+\varphi_{2}\right)\) If in the resultant wave the frequency and amplitude remain equal to those of superimposing waves. Then phase difference between them is
1 \(\dfrac{2 \pi}{3}\)
2 \(\dfrac{\pi}{3}\)
3 \(\dfrac{\pi}{6}\)
4 \(\dfrac{\pi}{4}\)
Explanation:
Superposition of waves does not alter the frequency of resultant wave and resultant amplitude \(\begin{aligned}& \Rightarrow a^{2}=a^{2}+a^{2}+2 a^{2} \cos \varphi=2 a^{2}(1+\cos \varphi) \\& \Rightarrow \cos \varphi=-1 / 2=\cos 2 \pi / 3 \therefore \varphi=2 \pi / 3 .\end{aligned}\)
PHXI15:WAVES
355094
A sonometer wire under suitable tension having specific gravity \(\rho\), vibrates with frequency \(n\) in air. If the load is completely immersed in water the frequency of vibration of wire will become
The fundamental frequency of sonometer wire, \(n=\dfrac{1}{2 L} \sqrt{\dfrac{T}{\mu}}\) As \(L\) and \(\mu\) are constant, so \(n\) depends on \(T\) \(\begin{aligned}& n \propto \sqrt{T} \\& \therefore \dfrac{n_{2}}{n_{1}}=\sqrt{\dfrac{T_{2}}{T_{1}}}=\sqrt{\dfrac{(\text { weight })_{s}-(\text { weight })_{w}}{(\text { weight })_{s}}} \\& =\sqrt{\dfrac{V \rho_{s} g-V \rho_{w} g}{V \rho_{s} g}}=\sqrt{\dfrac{\dfrac{\rho_{s}}{\rho_{w}}-1}{\dfrac{\rho_{s}}{\rho_{w}}}}=\sqrt{\dfrac{\rho-1}{\rho}}\end{aligned}\) \(\because \left( {{\text{specific}}\,{\text{gravity}}\,{\text{of}}\,{\text{sonometer}}{\mkern 1mu} \,{\text{wire}},{\text{ }}\rho {\text{ = }}\frac{{{\rho _s}}}{{{\rho _w}}}} \right)\) \(n_{2}=n \sqrt{\dfrac{\rho-1}{\rho}}\)
MHTCET - 2020
PHXI15:WAVES
355095
The length of a sonometer wire is \(90\;cm\) and the stationary wave setup in the wire is represented by an equation \(y=6 \sin \left(\dfrac{\pi x}{30}\right) \cos (250 t)\) where \(x, y\) are in \(\mathrm{cm}\) and \(t\) is in second. The number of loops is
1 1
2 2
3 4
4 3
Explanation:
Comparing the given equation with standard \(\frac{{2\pi }}{\lambda } = \frac{\pi }{{30}} \Rightarrow \lambda = 60\;cm\) \(\frac{\lambda }{2} = 30\;cm\) Since the length of sonometer wire is \(90 \mathrm{~cm}\) number of loops \(=\dfrac{90}{30}=3\).
355092
Two waves represented by \(y=a \sin (\omega t-k x)\) and \(y_{2}=\sqrt{3} a \cos (\omega t-k x)\) are superposed. The amplitude of the resultant wave is
1 \(2 a\)
2 \((2+\sqrt{3}) a\)
3 \((2-\sqrt{3}) a\)
4 \(2 \sqrt{3} a\)
Explanation:
Path difference \(=\pi / 2\) \(\therefore \quad a_{R}=\sqrt{a^{2}+(\sqrt{3} a)^{2}+2 \cdot a \cdot \sqrt{3} a \cdot \cos \pi / 2}=2 a\).
PHXI15:WAVES
355093
Two waves having equations \(x_{1}=a \sin \left(\omega t+\varphi_{1}\right), x_{2}=a \sin \left(\omega t+\varphi_{2}\right)\) If in the resultant wave the frequency and amplitude remain equal to those of superimposing waves. Then phase difference between them is
1 \(\dfrac{2 \pi}{3}\)
2 \(\dfrac{\pi}{3}\)
3 \(\dfrac{\pi}{6}\)
4 \(\dfrac{\pi}{4}\)
Explanation:
Superposition of waves does not alter the frequency of resultant wave and resultant amplitude \(\begin{aligned}& \Rightarrow a^{2}=a^{2}+a^{2}+2 a^{2} \cos \varphi=2 a^{2}(1+\cos \varphi) \\& \Rightarrow \cos \varphi=-1 / 2=\cos 2 \pi / 3 \therefore \varphi=2 \pi / 3 .\end{aligned}\)
PHXI15:WAVES
355094
A sonometer wire under suitable tension having specific gravity \(\rho\), vibrates with frequency \(n\) in air. If the load is completely immersed in water the frequency of vibration of wire will become
The fundamental frequency of sonometer wire, \(n=\dfrac{1}{2 L} \sqrt{\dfrac{T}{\mu}}\) As \(L\) and \(\mu\) are constant, so \(n\) depends on \(T\) \(\begin{aligned}& n \propto \sqrt{T} \\& \therefore \dfrac{n_{2}}{n_{1}}=\sqrt{\dfrac{T_{2}}{T_{1}}}=\sqrt{\dfrac{(\text { weight })_{s}-(\text { weight })_{w}}{(\text { weight })_{s}}} \\& =\sqrt{\dfrac{V \rho_{s} g-V \rho_{w} g}{V \rho_{s} g}}=\sqrt{\dfrac{\dfrac{\rho_{s}}{\rho_{w}}-1}{\dfrac{\rho_{s}}{\rho_{w}}}}=\sqrt{\dfrac{\rho-1}{\rho}}\end{aligned}\) \(\because \left( {{\text{specific}}\,{\text{gravity}}\,{\text{of}}\,{\text{sonometer}}{\mkern 1mu} \,{\text{wire}},{\text{ }}\rho {\text{ = }}\frac{{{\rho _s}}}{{{\rho _w}}}} \right)\) \(n_{2}=n \sqrt{\dfrac{\rho-1}{\rho}}\)
MHTCET - 2020
PHXI15:WAVES
355095
The length of a sonometer wire is \(90\;cm\) and the stationary wave setup in the wire is represented by an equation \(y=6 \sin \left(\dfrac{\pi x}{30}\right) \cos (250 t)\) where \(x, y\) are in \(\mathrm{cm}\) and \(t\) is in second. The number of loops is
1 1
2 2
3 4
4 3
Explanation:
Comparing the given equation with standard \(\frac{{2\pi }}{\lambda } = \frac{\pi }{{30}} \Rightarrow \lambda = 60\;cm\) \(\frac{\lambda }{2} = 30\;cm\) Since the length of sonometer wire is \(90 \mathrm{~cm}\) number of loops \(=\dfrac{90}{30}=3\).
355092
Two waves represented by \(y=a \sin (\omega t-k x)\) and \(y_{2}=\sqrt{3} a \cos (\omega t-k x)\) are superposed. The amplitude of the resultant wave is
1 \(2 a\)
2 \((2+\sqrt{3}) a\)
3 \((2-\sqrt{3}) a\)
4 \(2 \sqrt{3} a\)
Explanation:
Path difference \(=\pi / 2\) \(\therefore \quad a_{R}=\sqrt{a^{2}+(\sqrt{3} a)^{2}+2 \cdot a \cdot \sqrt{3} a \cdot \cos \pi / 2}=2 a\).
PHXI15:WAVES
355093
Two waves having equations \(x_{1}=a \sin \left(\omega t+\varphi_{1}\right), x_{2}=a \sin \left(\omega t+\varphi_{2}\right)\) If in the resultant wave the frequency and amplitude remain equal to those of superimposing waves. Then phase difference between them is
1 \(\dfrac{2 \pi}{3}\)
2 \(\dfrac{\pi}{3}\)
3 \(\dfrac{\pi}{6}\)
4 \(\dfrac{\pi}{4}\)
Explanation:
Superposition of waves does not alter the frequency of resultant wave and resultant amplitude \(\begin{aligned}& \Rightarrow a^{2}=a^{2}+a^{2}+2 a^{2} \cos \varphi=2 a^{2}(1+\cos \varphi) \\& \Rightarrow \cos \varphi=-1 / 2=\cos 2 \pi / 3 \therefore \varphi=2 \pi / 3 .\end{aligned}\)
PHXI15:WAVES
355094
A sonometer wire under suitable tension having specific gravity \(\rho\), vibrates with frequency \(n\) in air. If the load is completely immersed in water the frequency of vibration of wire will become
The fundamental frequency of sonometer wire, \(n=\dfrac{1}{2 L} \sqrt{\dfrac{T}{\mu}}\) As \(L\) and \(\mu\) are constant, so \(n\) depends on \(T\) \(\begin{aligned}& n \propto \sqrt{T} \\& \therefore \dfrac{n_{2}}{n_{1}}=\sqrt{\dfrac{T_{2}}{T_{1}}}=\sqrt{\dfrac{(\text { weight })_{s}-(\text { weight })_{w}}{(\text { weight })_{s}}} \\& =\sqrt{\dfrac{V \rho_{s} g-V \rho_{w} g}{V \rho_{s} g}}=\sqrt{\dfrac{\dfrac{\rho_{s}}{\rho_{w}}-1}{\dfrac{\rho_{s}}{\rho_{w}}}}=\sqrt{\dfrac{\rho-1}{\rho}}\end{aligned}\) \(\because \left( {{\text{specific}}\,{\text{gravity}}\,{\text{of}}\,{\text{sonometer}}{\mkern 1mu} \,{\text{wire}},{\text{ }}\rho {\text{ = }}\frac{{{\rho _s}}}{{{\rho _w}}}} \right)\) \(n_{2}=n \sqrt{\dfrac{\rho-1}{\rho}}\)
MHTCET - 2020
PHXI15:WAVES
355095
The length of a sonometer wire is \(90\;cm\) and the stationary wave setup in the wire is represented by an equation \(y=6 \sin \left(\dfrac{\pi x}{30}\right) \cos (250 t)\) where \(x, y\) are in \(\mathrm{cm}\) and \(t\) is in second. The number of loops is
1 1
2 2
3 4
4 3
Explanation:
Comparing the given equation with standard \(\frac{{2\pi }}{\lambda } = \frac{\pi }{{30}} \Rightarrow \lambda = 60\;cm\) \(\frac{\lambda }{2} = 30\;cm\) Since the length of sonometer wire is \(90 \mathrm{~cm}\) number of loops \(=\dfrac{90}{30}=3\).
355092
Two waves represented by \(y=a \sin (\omega t-k x)\) and \(y_{2}=\sqrt{3} a \cos (\omega t-k x)\) are superposed. The amplitude of the resultant wave is
1 \(2 a\)
2 \((2+\sqrt{3}) a\)
3 \((2-\sqrt{3}) a\)
4 \(2 \sqrt{3} a\)
Explanation:
Path difference \(=\pi / 2\) \(\therefore \quad a_{R}=\sqrt{a^{2}+(\sqrt{3} a)^{2}+2 \cdot a \cdot \sqrt{3} a \cdot \cos \pi / 2}=2 a\).
PHXI15:WAVES
355093
Two waves having equations \(x_{1}=a \sin \left(\omega t+\varphi_{1}\right), x_{2}=a \sin \left(\omega t+\varphi_{2}\right)\) If in the resultant wave the frequency and amplitude remain equal to those of superimposing waves. Then phase difference between them is
1 \(\dfrac{2 \pi}{3}\)
2 \(\dfrac{\pi}{3}\)
3 \(\dfrac{\pi}{6}\)
4 \(\dfrac{\pi}{4}\)
Explanation:
Superposition of waves does not alter the frequency of resultant wave and resultant amplitude \(\begin{aligned}& \Rightarrow a^{2}=a^{2}+a^{2}+2 a^{2} \cos \varphi=2 a^{2}(1+\cos \varphi) \\& \Rightarrow \cos \varphi=-1 / 2=\cos 2 \pi / 3 \therefore \varphi=2 \pi / 3 .\end{aligned}\)
PHXI15:WAVES
355094
A sonometer wire under suitable tension having specific gravity \(\rho\), vibrates with frequency \(n\) in air. If the load is completely immersed in water the frequency of vibration of wire will become
The fundamental frequency of sonometer wire, \(n=\dfrac{1}{2 L} \sqrt{\dfrac{T}{\mu}}\) As \(L\) and \(\mu\) are constant, so \(n\) depends on \(T\) \(\begin{aligned}& n \propto \sqrt{T} \\& \therefore \dfrac{n_{2}}{n_{1}}=\sqrt{\dfrac{T_{2}}{T_{1}}}=\sqrt{\dfrac{(\text { weight })_{s}-(\text { weight })_{w}}{(\text { weight })_{s}}} \\& =\sqrt{\dfrac{V \rho_{s} g-V \rho_{w} g}{V \rho_{s} g}}=\sqrt{\dfrac{\dfrac{\rho_{s}}{\rho_{w}}-1}{\dfrac{\rho_{s}}{\rho_{w}}}}=\sqrt{\dfrac{\rho-1}{\rho}}\end{aligned}\) \(\because \left( {{\text{specific}}\,{\text{gravity}}\,{\text{of}}\,{\text{sonometer}}{\mkern 1mu} \,{\text{wire}},{\text{ }}\rho {\text{ = }}\frac{{{\rho _s}}}{{{\rho _w}}}} \right)\) \(n_{2}=n \sqrt{\dfrac{\rho-1}{\rho}}\)
MHTCET - 2020
PHXI15:WAVES
355095
The length of a sonometer wire is \(90\;cm\) and the stationary wave setup in the wire is represented by an equation \(y=6 \sin \left(\dfrac{\pi x}{30}\right) \cos (250 t)\) where \(x, y\) are in \(\mathrm{cm}\) and \(t\) is in second. The number of loops is
1 1
2 2
3 4
4 3
Explanation:
Comparing the given equation with standard \(\frac{{2\pi }}{\lambda } = \frac{\pi }{{30}} \Rightarrow \lambda = 60\;cm\) \(\frac{\lambda }{2} = 30\;cm\) Since the length of sonometer wire is \(90 \mathrm{~cm}\) number of loops \(=\dfrac{90}{30}=3\).
