354575
Sound waves of wavelength \(\lambda\) travelling in a medium with a speed of \(vm/s\) enter into another medium where its speed is \(2vm/s\). Wavelength of sound waves in the second medium is
1 \(\lambda\)
2 \(\dfrac{\lambda}{2}\)
3 \(2 \lambda\)
4 \(4 \lambda\)
Explanation:
In the first medium, frequency, \(f=\dfrac{v}{\lambda}\) Frequency remains the same in second medium, i.e., \(f=f^{\prime}\) \(\therefore \dfrac{v^{\prime}}{\lambda^{\prime}}=\dfrac{2 v}{\lambda^{\prime}}=\dfrac{v}{\lambda}\) \(\therefore \quad \lambda^{\prime}=2 \lambda\)
NCERT Exemplar
PHXI15:WAVES
354576
In sinusoidal wave, the time required for a particular point to move from maximum displacement to zero displacement is \(0 \cdot 14\) second. The frequency of the wave is
354577
\(y(x, t)=\dfrac{1.0}{\left[(a x+b t)^{2}+(c x+d t)^{2}+5\right]} s\) represents a moving pulse where \(x\) and \(y\) are in metres and \(t\) in seconds. It has speed of pulse \(\dfrac{b}{a}\). Find the correct option.
1 \(b c=a d\)
2 \(b d=a c\)
3 \(a b=c d\)
4 \(b d=-a c\)
Explanation:
Speed of pulse \(=\dfrac{b}{a}=\dfrac{d}{c}\).
PHXI15:WAVES
354578
The equation of a wave is given as \(y=0.07 \sin (12 \pi x-3000 \pi t)\). where \(x\) is in metre and \(t\) in sec, then the correct statement is
1 \(\lambda = 1/6m,v = 250\;m/s\)
2 \(a = 0.07\;m,v = 300\;m/s\)
3 \(n = 1500,v = 200\;m/s\)
4 None of these
Explanation:
Comparing the given equation with \(y=a \sin (\omega t-k x)\) We get \(\omega=3000 \pi \Rightarrow n=\dfrac{\omega}{2 \pi}=1500 H z\) and \(k=\dfrac{2 \pi}{\lambda}=12 \pi \Rightarrow \lambda=\dfrac{1}{6} m\) So, \(v = n\lambda \Rightarrow v = 1500 \times \frac{1}{6} = 250\;m/s\)
354575
Sound waves of wavelength \(\lambda\) travelling in a medium with a speed of \(vm/s\) enter into another medium where its speed is \(2vm/s\). Wavelength of sound waves in the second medium is
1 \(\lambda\)
2 \(\dfrac{\lambda}{2}\)
3 \(2 \lambda\)
4 \(4 \lambda\)
Explanation:
In the first medium, frequency, \(f=\dfrac{v}{\lambda}\) Frequency remains the same in second medium, i.e., \(f=f^{\prime}\) \(\therefore \dfrac{v^{\prime}}{\lambda^{\prime}}=\dfrac{2 v}{\lambda^{\prime}}=\dfrac{v}{\lambda}\) \(\therefore \quad \lambda^{\prime}=2 \lambda\)
NCERT Exemplar
PHXI15:WAVES
354576
In sinusoidal wave, the time required for a particular point to move from maximum displacement to zero displacement is \(0 \cdot 14\) second. The frequency of the wave is
354577
\(y(x, t)=\dfrac{1.0}{\left[(a x+b t)^{2}+(c x+d t)^{2}+5\right]} s\) represents a moving pulse where \(x\) and \(y\) are in metres and \(t\) in seconds. It has speed of pulse \(\dfrac{b}{a}\). Find the correct option.
1 \(b c=a d\)
2 \(b d=a c\)
3 \(a b=c d\)
4 \(b d=-a c\)
Explanation:
Speed of pulse \(=\dfrac{b}{a}=\dfrac{d}{c}\).
PHXI15:WAVES
354578
The equation of a wave is given as \(y=0.07 \sin (12 \pi x-3000 \pi t)\). where \(x\) is in metre and \(t\) in sec, then the correct statement is
1 \(\lambda = 1/6m,v = 250\;m/s\)
2 \(a = 0.07\;m,v = 300\;m/s\)
3 \(n = 1500,v = 200\;m/s\)
4 None of these
Explanation:
Comparing the given equation with \(y=a \sin (\omega t-k x)\) We get \(\omega=3000 \pi \Rightarrow n=\dfrac{\omega}{2 \pi}=1500 H z\) and \(k=\dfrac{2 \pi}{\lambda}=12 \pi \Rightarrow \lambda=\dfrac{1}{6} m\) So, \(v = n\lambda \Rightarrow v = 1500 \times \frac{1}{6} = 250\;m/s\)
354575
Sound waves of wavelength \(\lambda\) travelling in a medium with a speed of \(vm/s\) enter into another medium where its speed is \(2vm/s\). Wavelength of sound waves in the second medium is
1 \(\lambda\)
2 \(\dfrac{\lambda}{2}\)
3 \(2 \lambda\)
4 \(4 \lambda\)
Explanation:
In the first medium, frequency, \(f=\dfrac{v}{\lambda}\) Frequency remains the same in second medium, i.e., \(f=f^{\prime}\) \(\therefore \dfrac{v^{\prime}}{\lambda^{\prime}}=\dfrac{2 v}{\lambda^{\prime}}=\dfrac{v}{\lambda}\) \(\therefore \quad \lambda^{\prime}=2 \lambda\)
NCERT Exemplar
PHXI15:WAVES
354576
In sinusoidal wave, the time required for a particular point to move from maximum displacement to zero displacement is \(0 \cdot 14\) second. The frequency of the wave is
354577
\(y(x, t)=\dfrac{1.0}{\left[(a x+b t)^{2}+(c x+d t)^{2}+5\right]} s\) represents a moving pulse where \(x\) and \(y\) are in metres and \(t\) in seconds. It has speed of pulse \(\dfrac{b}{a}\). Find the correct option.
1 \(b c=a d\)
2 \(b d=a c\)
3 \(a b=c d\)
4 \(b d=-a c\)
Explanation:
Speed of pulse \(=\dfrac{b}{a}=\dfrac{d}{c}\).
PHXI15:WAVES
354578
The equation of a wave is given as \(y=0.07 \sin (12 \pi x-3000 \pi t)\). where \(x\) is in metre and \(t\) in sec, then the correct statement is
1 \(\lambda = 1/6m,v = 250\;m/s\)
2 \(a = 0.07\;m,v = 300\;m/s\)
3 \(n = 1500,v = 200\;m/s\)
4 None of these
Explanation:
Comparing the given equation with \(y=a \sin (\omega t-k x)\) We get \(\omega=3000 \pi \Rightarrow n=\dfrac{\omega}{2 \pi}=1500 H z\) and \(k=\dfrac{2 \pi}{\lambda}=12 \pi \Rightarrow \lambda=\dfrac{1}{6} m\) So, \(v = n\lambda \Rightarrow v = 1500 \times \frac{1}{6} = 250\;m/s\)
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PHXI15:WAVES
354575
Sound waves of wavelength \(\lambda\) travelling in a medium with a speed of \(vm/s\) enter into another medium where its speed is \(2vm/s\). Wavelength of sound waves in the second medium is
1 \(\lambda\)
2 \(\dfrac{\lambda}{2}\)
3 \(2 \lambda\)
4 \(4 \lambda\)
Explanation:
In the first medium, frequency, \(f=\dfrac{v}{\lambda}\) Frequency remains the same in second medium, i.e., \(f=f^{\prime}\) \(\therefore \dfrac{v^{\prime}}{\lambda^{\prime}}=\dfrac{2 v}{\lambda^{\prime}}=\dfrac{v}{\lambda}\) \(\therefore \quad \lambda^{\prime}=2 \lambda\)
NCERT Exemplar
PHXI15:WAVES
354576
In sinusoidal wave, the time required for a particular point to move from maximum displacement to zero displacement is \(0 \cdot 14\) second. The frequency of the wave is
354577
\(y(x, t)=\dfrac{1.0}{\left[(a x+b t)^{2}+(c x+d t)^{2}+5\right]} s\) represents a moving pulse where \(x\) and \(y\) are in metres and \(t\) in seconds. It has speed of pulse \(\dfrac{b}{a}\). Find the correct option.
1 \(b c=a d\)
2 \(b d=a c\)
3 \(a b=c d\)
4 \(b d=-a c\)
Explanation:
Speed of pulse \(=\dfrac{b}{a}=\dfrac{d}{c}\).
PHXI15:WAVES
354578
The equation of a wave is given as \(y=0.07 \sin (12 \pi x-3000 \pi t)\). where \(x\) is in metre and \(t\) in sec, then the correct statement is
1 \(\lambda = 1/6m,v = 250\;m/s\)
2 \(a = 0.07\;m,v = 300\;m/s\)
3 \(n = 1500,v = 200\;m/s\)
4 None of these
Explanation:
Comparing the given equation with \(y=a \sin (\omega t-k x)\) We get \(\omega=3000 \pi \Rightarrow n=\dfrac{\omega}{2 \pi}=1500 H z\) and \(k=\dfrac{2 \pi}{\lambda}=12 \pi \Rightarrow \lambda=\dfrac{1}{6} m\) So, \(v = n\lambda \Rightarrow v = 1500 \times \frac{1}{6} = 250\;m/s\)