355190
In order to double the frequency of the fundamental note emitted by a stretched string, the length is reduced to \(\dfrac{3}{4}\) \(th\) of the original length and the tension is changed. The factor by which the tension is to be changed is
1 \(\dfrac{8}{9}\)
2 \(\dfrac{3}{8}\)
3 \(\dfrac{9}{4}\)
4 \(\dfrac{2}{3}\)
Explanation:
\(f=\dfrac{1}{2 l} \sqrt{\dfrac{T}{\mu}} \Rightarrow f \propto \dfrac{\sqrt{T}}{l}\) \(\Rightarrow \dfrac{T_{2}}{T_{1}}=\left(\dfrac{f_{2}}{f_{1}}\right)^{2}\left(\dfrac{l_{2}}{l_{1}}\right)^{2}=(2)^{2}\left(\dfrac{3}{4}\right)^{2}=\dfrac{9}{4}\)
PHXI15:WAVES
355191
A stretched string of length \(l\), fixed at both ends can sustain stationary waves of wavelength \(\lambda\), given by
1 \(\lambda=\dfrac{2 l}{n}\)
2 \(\lambda=2 \ln\)
3 \(\lambda=\dfrac{\mathrm{n}^{2}}{2 l}\)
4 \(\lambda=\dfrac{l^{2}}{2 n}\)
Explanation:
Conceptual Question
PHXI15:WAVES
355192
A string vibrates with a frequency of \(200\;Hz\). When its length is doubled and tension is altered, it begins to vibrate with a frequency of \(300\;Hz.\) The ratio of the new tension to the original tension is
1 \(3: 1\)
2 \(1: 3\)
3 \(9: 1\)
4 \(1: 9\)
Explanation:
Initial frequency \(f_{1}=\dfrac{1}{2 l} \sqrt{\dfrac{T_{1}}{\mu}} \Rightarrow T_{1}=f_{1}^{2} \cdot 4 l^{2} \cdot \mu\) When length of string is doubled, \(T_{2}=f_{2}^{2} \cdot 4 \times 4 l^{2} \cdot \mu\) Given \({f_1} = 200\;Hz\) and \({f_2} = 300\;Hz\), \(\therefore \quad \dfrac{T_{2}}{T_{1}}=\dfrac{(300)^{2}}{(200)^{2}} \times 4=\dfrac{9}{1}\)
PHXI15:WAVES
355193
In a stationary wave represented by \(y=2 a \cos (k x) \sin (\omega t)\). The intensity at a certain point is maximum when:
1 \(\cos k x\) is minimum
2 \(\cos k x\) is maximum
3 \(\sin \omega t\) is minimum
4 \(\sin \omega t\) is maximum
Explanation:
In stationary wave \(y=2 a \cos k x \sin \omega t\) Amplitude of the stationary wave is \(A=2 a \cos (k x)\) Intensity at certain point is maximum when amplitude is maximum. Intensity is maximum when \(\cos (k x)\) is maximum
PHXI15:WAVES
355194
For stationary wave, \(y=10 \sin (300 \pi t) \sin \left(\dfrac{\pi}{8} x\right)\) The distance between consecutive nodes is
355190
In order to double the frequency of the fundamental note emitted by a stretched string, the length is reduced to \(\dfrac{3}{4}\) \(th\) of the original length and the tension is changed. The factor by which the tension is to be changed is
1 \(\dfrac{8}{9}\)
2 \(\dfrac{3}{8}\)
3 \(\dfrac{9}{4}\)
4 \(\dfrac{2}{3}\)
Explanation:
\(f=\dfrac{1}{2 l} \sqrt{\dfrac{T}{\mu}} \Rightarrow f \propto \dfrac{\sqrt{T}}{l}\) \(\Rightarrow \dfrac{T_{2}}{T_{1}}=\left(\dfrac{f_{2}}{f_{1}}\right)^{2}\left(\dfrac{l_{2}}{l_{1}}\right)^{2}=(2)^{2}\left(\dfrac{3}{4}\right)^{2}=\dfrac{9}{4}\)
PHXI15:WAVES
355191
A stretched string of length \(l\), fixed at both ends can sustain stationary waves of wavelength \(\lambda\), given by
1 \(\lambda=\dfrac{2 l}{n}\)
2 \(\lambda=2 \ln\)
3 \(\lambda=\dfrac{\mathrm{n}^{2}}{2 l}\)
4 \(\lambda=\dfrac{l^{2}}{2 n}\)
Explanation:
Conceptual Question
PHXI15:WAVES
355192
A string vibrates with a frequency of \(200\;Hz\). When its length is doubled and tension is altered, it begins to vibrate with a frequency of \(300\;Hz.\) The ratio of the new tension to the original tension is
1 \(3: 1\)
2 \(1: 3\)
3 \(9: 1\)
4 \(1: 9\)
Explanation:
Initial frequency \(f_{1}=\dfrac{1}{2 l} \sqrt{\dfrac{T_{1}}{\mu}} \Rightarrow T_{1}=f_{1}^{2} \cdot 4 l^{2} \cdot \mu\) When length of string is doubled, \(T_{2}=f_{2}^{2} \cdot 4 \times 4 l^{2} \cdot \mu\) Given \({f_1} = 200\;Hz\) and \({f_2} = 300\;Hz\), \(\therefore \quad \dfrac{T_{2}}{T_{1}}=\dfrac{(300)^{2}}{(200)^{2}} \times 4=\dfrac{9}{1}\)
PHXI15:WAVES
355193
In a stationary wave represented by \(y=2 a \cos (k x) \sin (\omega t)\). The intensity at a certain point is maximum when:
1 \(\cos k x\) is minimum
2 \(\cos k x\) is maximum
3 \(\sin \omega t\) is minimum
4 \(\sin \omega t\) is maximum
Explanation:
In stationary wave \(y=2 a \cos k x \sin \omega t\) Amplitude of the stationary wave is \(A=2 a \cos (k x)\) Intensity at certain point is maximum when amplitude is maximum. Intensity is maximum when \(\cos (k x)\) is maximum
PHXI15:WAVES
355194
For stationary wave, \(y=10 \sin (300 \pi t) \sin \left(\dfrac{\pi}{8} x\right)\) The distance between consecutive nodes is
355190
In order to double the frequency of the fundamental note emitted by a stretched string, the length is reduced to \(\dfrac{3}{4}\) \(th\) of the original length and the tension is changed. The factor by which the tension is to be changed is
1 \(\dfrac{8}{9}\)
2 \(\dfrac{3}{8}\)
3 \(\dfrac{9}{4}\)
4 \(\dfrac{2}{3}\)
Explanation:
\(f=\dfrac{1}{2 l} \sqrt{\dfrac{T}{\mu}} \Rightarrow f \propto \dfrac{\sqrt{T}}{l}\) \(\Rightarrow \dfrac{T_{2}}{T_{1}}=\left(\dfrac{f_{2}}{f_{1}}\right)^{2}\left(\dfrac{l_{2}}{l_{1}}\right)^{2}=(2)^{2}\left(\dfrac{3}{4}\right)^{2}=\dfrac{9}{4}\)
PHXI15:WAVES
355191
A stretched string of length \(l\), fixed at both ends can sustain stationary waves of wavelength \(\lambda\), given by
1 \(\lambda=\dfrac{2 l}{n}\)
2 \(\lambda=2 \ln\)
3 \(\lambda=\dfrac{\mathrm{n}^{2}}{2 l}\)
4 \(\lambda=\dfrac{l^{2}}{2 n}\)
Explanation:
Conceptual Question
PHXI15:WAVES
355192
A string vibrates with a frequency of \(200\;Hz\). When its length is doubled and tension is altered, it begins to vibrate with a frequency of \(300\;Hz.\) The ratio of the new tension to the original tension is
1 \(3: 1\)
2 \(1: 3\)
3 \(9: 1\)
4 \(1: 9\)
Explanation:
Initial frequency \(f_{1}=\dfrac{1}{2 l} \sqrt{\dfrac{T_{1}}{\mu}} \Rightarrow T_{1}=f_{1}^{2} \cdot 4 l^{2} \cdot \mu\) When length of string is doubled, \(T_{2}=f_{2}^{2} \cdot 4 \times 4 l^{2} \cdot \mu\) Given \({f_1} = 200\;Hz\) and \({f_2} = 300\;Hz\), \(\therefore \quad \dfrac{T_{2}}{T_{1}}=\dfrac{(300)^{2}}{(200)^{2}} \times 4=\dfrac{9}{1}\)
PHXI15:WAVES
355193
In a stationary wave represented by \(y=2 a \cos (k x) \sin (\omega t)\). The intensity at a certain point is maximum when:
1 \(\cos k x\) is minimum
2 \(\cos k x\) is maximum
3 \(\sin \omega t\) is minimum
4 \(\sin \omega t\) is maximum
Explanation:
In stationary wave \(y=2 a \cos k x \sin \omega t\) Amplitude of the stationary wave is \(A=2 a \cos (k x)\) Intensity at certain point is maximum when amplitude is maximum. Intensity is maximum when \(\cos (k x)\) is maximum
PHXI15:WAVES
355194
For stationary wave, \(y=10 \sin (300 \pi t) \sin \left(\dfrac{\pi}{8} x\right)\) The distance between consecutive nodes is
355190
In order to double the frequency of the fundamental note emitted by a stretched string, the length is reduced to \(\dfrac{3}{4}\) \(th\) of the original length and the tension is changed. The factor by which the tension is to be changed is
1 \(\dfrac{8}{9}\)
2 \(\dfrac{3}{8}\)
3 \(\dfrac{9}{4}\)
4 \(\dfrac{2}{3}\)
Explanation:
\(f=\dfrac{1}{2 l} \sqrt{\dfrac{T}{\mu}} \Rightarrow f \propto \dfrac{\sqrt{T}}{l}\) \(\Rightarrow \dfrac{T_{2}}{T_{1}}=\left(\dfrac{f_{2}}{f_{1}}\right)^{2}\left(\dfrac{l_{2}}{l_{1}}\right)^{2}=(2)^{2}\left(\dfrac{3}{4}\right)^{2}=\dfrac{9}{4}\)
PHXI15:WAVES
355191
A stretched string of length \(l\), fixed at both ends can sustain stationary waves of wavelength \(\lambda\), given by
1 \(\lambda=\dfrac{2 l}{n}\)
2 \(\lambda=2 \ln\)
3 \(\lambda=\dfrac{\mathrm{n}^{2}}{2 l}\)
4 \(\lambda=\dfrac{l^{2}}{2 n}\)
Explanation:
Conceptual Question
PHXI15:WAVES
355192
A string vibrates with a frequency of \(200\;Hz\). When its length is doubled and tension is altered, it begins to vibrate with a frequency of \(300\;Hz.\) The ratio of the new tension to the original tension is
1 \(3: 1\)
2 \(1: 3\)
3 \(9: 1\)
4 \(1: 9\)
Explanation:
Initial frequency \(f_{1}=\dfrac{1}{2 l} \sqrt{\dfrac{T_{1}}{\mu}} \Rightarrow T_{1}=f_{1}^{2} \cdot 4 l^{2} \cdot \mu\) When length of string is doubled, \(T_{2}=f_{2}^{2} \cdot 4 \times 4 l^{2} \cdot \mu\) Given \({f_1} = 200\;Hz\) and \({f_2} = 300\;Hz\), \(\therefore \quad \dfrac{T_{2}}{T_{1}}=\dfrac{(300)^{2}}{(200)^{2}} \times 4=\dfrac{9}{1}\)
PHXI15:WAVES
355193
In a stationary wave represented by \(y=2 a \cos (k x) \sin (\omega t)\). The intensity at a certain point is maximum when:
1 \(\cos k x\) is minimum
2 \(\cos k x\) is maximum
3 \(\sin \omega t\) is minimum
4 \(\sin \omega t\) is maximum
Explanation:
In stationary wave \(y=2 a \cos k x \sin \omega t\) Amplitude of the stationary wave is \(A=2 a \cos (k x)\) Intensity at certain point is maximum when amplitude is maximum. Intensity is maximum when \(\cos (k x)\) is maximum
PHXI15:WAVES
355194
For stationary wave, \(y=10 \sin (300 \pi t) \sin \left(\dfrac{\pi}{8} x\right)\) The distance between consecutive nodes is
355190
In order to double the frequency of the fundamental note emitted by a stretched string, the length is reduced to \(\dfrac{3}{4}\) \(th\) of the original length and the tension is changed. The factor by which the tension is to be changed is
1 \(\dfrac{8}{9}\)
2 \(\dfrac{3}{8}\)
3 \(\dfrac{9}{4}\)
4 \(\dfrac{2}{3}\)
Explanation:
\(f=\dfrac{1}{2 l} \sqrt{\dfrac{T}{\mu}} \Rightarrow f \propto \dfrac{\sqrt{T}}{l}\) \(\Rightarrow \dfrac{T_{2}}{T_{1}}=\left(\dfrac{f_{2}}{f_{1}}\right)^{2}\left(\dfrac{l_{2}}{l_{1}}\right)^{2}=(2)^{2}\left(\dfrac{3}{4}\right)^{2}=\dfrac{9}{4}\)
PHXI15:WAVES
355191
A stretched string of length \(l\), fixed at both ends can sustain stationary waves of wavelength \(\lambda\), given by
1 \(\lambda=\dfrac{2 l}{n}\)
2 \(\lambda=2 \ln\)
3 \(\lambda=\dfrac{\mathrm{n}^{2}}{2 l}\)
4 \(\lambda=\dfrac{l^{2}}{2 n}\)
Explanation:
Conceptual Question
PHXI15:WAVES
355192
A string vibrates with a frequency of \(200\;Hz\). When its length is doubled and tension is altered, it begins to vibrate with a frequency of \(300\;Hz.\) The ratio of the new tension to the original tension is
1 \(3: 1\)
2 \(1: 3\)
3 \(9: 1\)
4 \(1: 9\)
Explanation:
Initial frequency \(f_{1}=\dfrac{1}{2 l} \sqrt{\dfrac{T_{1}}{\mu}} \Rightarrow T_{1}=f_{1}^{2} \cdot 4 l^{2} \cdot \mu\) When length of string is doubled, \(T_{2}=f_{2}^{2} \cdot 4 \times 4 l^{2} \cdot \mu\) Given \({f_1} = 200\;Hz\) and \({f_2} = 300\;Hz\), \(\therefore \quad \dfrac{T_{2}}{T_{1}}=\dfrac{(300)^{2}}{(200)^{2}} \times 4=\dfrac{9}{1}\)
PHXI15:WAVES
355193
In a stationary wave represented by \(y=2 a \cos (k x) \sin (\omega t)\). The intensity at a certain point is maximum when:
1 \(\cos k x\) is minimum
2 \(\cos k x\) is maximum
3 \(\sin \omega t\) is minimum
4 \(\sin \omega t\) is maximum
Explanation:
In stationary wave \(y=2 a \cos k x \sin \omega t\) Amplitude of the stationary wave is \(A=2 a \cos (k x)\) Intensity at certain point is maximum when amplitude is maximum. Intensity is maximum when \(\cos (k x)\) is maximum
PHXI15:WAVES
355194
For stationary wave, \(y=10 \sin (300 \pi t) \sin \left(\dfrac{\pi}{8} x\right)\) The distance between consecutive nodes is
