355165
When a string fixed at both ends vibrates in 1 loop, 2 loops, 3 loops and 4 loops, the frequenices are in the ratio
1 \(1: 2: 3: 4\)
2 \(1: 1: 1: 1\)
3 \(1: 4: 9: 16\)
4 \(4: 3: 2: 1\)
Explanation:
When the string vibrates in \(n\) loops, its frequency is \(f_{n}=\dfrac{n v}{2 L}\) Where \(L\) is the length of the string and \(v\) is the velocity of the wave. \(\therefore\) When the string fixed at its both ends vibrates in 1 loop, 2 loops, 3 loops and 4 loops, the frequencies are in the ratio \(1: 2: 3: 4\)
PHXI15:WAVES
355166
A \(10\;cm\) long rubber band obeys Hooke's law. When the rubber band is stretched to a total length of \(12\;cm\) the lowest resonant frequency is \({f_0}\). The rubber band is then stretched to a length of \(13\;cm\). The lowest resonant frequency will now be
1 Lower than \(f_{0}\)
2 Higher than \(f_{0}\)
3 The same as \(f_{0}\)
4 Changed, but the change (positive or negative) depends on the elastic constant and the original tension
Explanation:
\(f_{0}=\dfrac{1}{2 l} \sqrt{\dfrac{T}{\mu}}\) For elastic wire \(T = Kx\) \(x \rightarrow\) elongation from natural length Hence frequency depends on \(K\) and original tension
PHXI15:WAVES
355167
For a stationary wave, \(Y = 10\sin \left( {\frac{{\pi x}}{{15}}} \right)\cos (48\pi t)cm\), the distance between a node and the successive antinode is
1 \(60\;cm\)
2 \(30\;cm\)
3 \(7.5\;cm\)
4 \(15\;cm\)
Explanation:
On comparing the given equation with a standard stationary wave equation, we get \(k = \frac{\pi }{{15}} = \frac{{2\pi }}{\lambda } \Rightarrow \lambda = 30\;cm\) Distance between node and antinode is, \(\frac{\lambda }{4} = \frac{{30}}{4} = 7.5\;cm\)
355165
When a string fixed at both ends vibrates in 1 loop, 2 loops, 3 loops and 4 loops, the frequenices are in the ratio
1 \(1: 2: 3: 4\)
2 \(1: 1: 1: 1\)
3 \(1: 4: 9: 16\)
4 \(4: 3: 2: 1\)
Explanation:
When the string vibrates in \(n\) loops, its frequency is \(f_{n}=\dfrac{n v}{2 L}\) Where \(L\) is the length of the string and \(v\) is the velocity of the wave. \(\therefore\) When the string fixed at its both ends vibrates in 1 loop, 2 loops, 3 loops and 4 loops, the frequencies are in the ratio \(1: 2: 3: 4\)
PHXI15:WAVES
355166
A \(10\;cm\) long rubber band obeys Hooke's law. When the rubber band is stretched to a total length of \(12\;cm\) the lowest resonant frequency is \({f_0}\). The rubber band is then stretched to a length of \(13\;cm\). The lowest resonant frequency will now be
1 Lower than \(f_{0}\)
2 Higher than \(f_{0}\)
3 The same as \(f_{0}\)
4 Changed, but the change (positive or negative) depends on the elastic constant and the original tension
Explanation:
\(f_{0}=\dfrac{1}{2 l} \sqrt{\dfrac{T}{\mu}}\) For elastic wire \(T = Kx\) \(x \rightarrow\) elongation from natural length Hence frequency depends on \(K\) and original tension
PHXI15:WAVES
355167
For a stationary wave, \(Y = 10\sin \left( {\frac{{\pi x}}{{15}}} \right)\cos (48\pi t)cm\), the distance between a node and the successive antinode is
1 \(60\;cm\)
2 \(30\;cm\)
3 \(7.5\;cm\)
4 \(15\;cm\)
Explanation:
On comparing the given equation with a standard stationary wave equation, we get \(k = \frac{\pi }{{15}} = \frac{{2\pi }}{\lambda } \Rightarrow \lambda = 30\;cm\) Distance between node and antinode is, \(\frac{\lambda }{4} = \frac{{30}}{4} = 7.5\;cm\)
355165
When a string fixed at both ends vibrates in 1 loop, 2 loops, 3 loops and 4 loops, the frequenices are in the ratio
1 \(1: 2: 3: 4\)
2 \(1: 1: 1: 1\)
3 \(1: 4: 9: 16\)
4 \(4: 3: 2: 1\)
Explanation:
When the string vibrates in \(n\) loops, its frequency is \(f_{n}=\dfrac{n v}{2 L}\) Where \(L\) is the length of the string and \(v\) is the velocity of the wave. \(\therefore\) When the string fixed at its both ends vibrates in 1 loop, 2 loops, 3 loops and 4 loops, the frequencies are in the ratio \(1: 2: 3: 4\)
PHXI15:WAVES
355166
A \(10\;cm\) long rubber band obeys Hooke's law. When the rubber band is stretched to a total length of \(12\;cm\) the lowest resonant frequency is \({f_0}\). The rubber band is then stretched to a length of \(13\;cm\). The lowest resonant frequency will now be
1 Lower than \(f_{0}\)
2 Higher than \(f_{0}\)
3 The same as \(f_{0}\)
4 Changed, but the change (positive or negative) depends on the elastic constant and the original tension
Explanation:
\(f_{0}=\dfrac{1}{2 l} \sqrt{\dfrac{T}{\mu}}\) For elastic wire \(T = Kx\) \(x \rightarrow\) elongation from natural length Hence frequency depends on \(K\) and original tension
PHXI15:WAVES
355167
For a stationary wave, \(Y = 10\sin \left( {\frac{{\pi x}}{{15}}} \right)\cos (48\pi t)cm\), the distance between a node and the successive antinode is
1 \(60\;cm\)
2 \(30\;cm\)
3 \(7.5\;cm\)
4 \(15\;cm\)
Explanation:
On comparing the given equation with a standard stationary wave equation, we get \(k = \frac{\pi }{{15}} = \frac{{2\pi }}{\lambda } \Rightarrow \lambda = 30\;cm\) Distance between node and antinode is, \(\frac{\lambda }{4} = \frac{{30}}{4} = 7.5\;cm\)
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PHXI15:WAVES
355165
When a string fixed at both ends vibrates in 1 loop, 2 loops, 3 loops and 4 loops, the frequenices are in the ratio
1 \(1: 2: 3: 4\)
2 \(1: 1: 1: 1\)
3 \(1: 4: 9: 16\)
4 \(4: 3: 2: 1\)
Explanation:
When the string vibrates in \(n\) loops, its frequency is \(f_{n}=\dfrac{n v}{2 L}\) Where \(L\) is the length of the string and \(v\) is the velocity of the wave. \(\therefore\) When the string fixed at its both ends vibrates in 1 loop, 2 loops, 3 loops and 4 loops, the frequencies are in the ratio \(1: 2: 3: 4\)
PHXI15:WAVES
355166
A \(10\;cm\) long rubber band obeys Hooke's law. When the rubber band is stretched to a total length of \(12\;cm\) the lowest resonant frequency is \({f_0}\). The rubber band is then stretched to a length of \(13\;cm\). The lowest resonant frequency will now be
1 Lower than \(f_{0}\)
2 Higher than \(f_{0}\)
3 The same as \(f_{0}\)
4 Changed, but the change (positive or negative) depends on the elastic constant and the original tension
Explanation:
\(f_{0}=\dfrac{1}{2 l} \sqrt{\dfrac{T}{\mu}}\) For elastic wire \(T = Kx\) \(x \rightarrow\) elongation from natural length Hence frequency depends on \(K\) and original tension
PHXI15:WAVES
355167
For a stationary wave, \(Y = 10\sin \left( {\frac{{\pi x}}{{15}}} \right)\cos (48\pi t)cm\), the distance between a node and the successive antinode is
1 \(60\;cm\)
2 \(30\;cm\)
3 \(7.5\;cm\)
4 \(15\;cm\)
Explanation:
On comparing the given equation with a standard stationary wave equation, we get \(k = \frac{\pi }{{15}} = \frac{{2\pi }}{\lambda } \Rightarrow \lambda = 30\;cm\) Distance between node and antinode is, \(\frac{\lambda }{4} = \frac{{30}}{4} = 7.5\;cm\)
