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PHXI15:WAVES

355169 In a stationary wave

1 There is no net transfer of energy

2 Energy is constant at all points

3 Amplitude is same for all points

4 Energy and amplitude is same at all points

PHXI15:WAVES

355170 The equation of a stationary wave is $y = 2 \sin (\frac{π x}{15}) \cos (48 π t)$ . The distance between a node and its next antinode is

1 22.5 units

2 7.5 units

3 30 units

4 1.5 units

Explanation:

Comparing the given equation with standard equation of stationary wave
$y = A \sin (k x) \cos (ω t)$ ,
$k = \frac{2 π}{λ} = \frac{π}{15} \Rightarrow λ = 30$
Distance between a node and next antinode $= \frac{λ}{4} = \frac{30}{4} = 7.5$ units

KCET - 2019

PHXI15:WAVES

355171 A string vibrates with a frequency 200 $H z$ . Its length is doubled and its tension is altered until it begins to vibrate with frequency 300 $H z$ . What is the ratio of the new tension to the original tension?

1 3

2 9

3 12

4 5

Explanation:

Let $f_{1}$ and $f_{2}$ be the frequencies in two cases and $L_{1}, L_{2}$ and $T_{1}, T_{2}$ be respectively the corresponding lengths of the string and tensions under which it is vibrating.
$f_{1} = \frac{1}{2 L_{1}} \sqrt{\frac{T_{1}}{μ}}$
and $f_{2} = \frac{1}{2 L_{2}} \sqrt{\frac{T_{2}}{μ}}$
Thus, $\frac{f_{1}}{f_{2}} = \frac{L_{2}}{L_{1}} \sqrt{\frac{T_{1}}{T_{2}}}$ or $\frac{200}{300} = 2 \sqrt{\frac{T_{1}}{T_{2}}}$ $(f_{1} = 200, f_{2} = 300$ and $L_{2} = 2 L_{1}$ or $L_{2} / L_{1} = 2)$
or $\frac{2}{3} = 2 \sqrt{\frac{T_{1}}{T_{2}}}$ or $\frac{4}{9} = 4 \frac{T_{1}}{T_{2}}$ or $\frac{T_{2}}{T_{1}} = 9$

PHXI15:WAVES

355172 Two vibrating strings of the same material but lengths $L$ and $2 L$ have radii $2 r$ and $r$ respectively. They are stretched under the same tension. Both the strings vibrate in their fundamental modes, the one of length $L$ with frequency $v_{1}$ and the other with frequency $v_{2}$ . The ratio $v_{1} / v_{2}$ is

1 2

2 4

3 3

4 1

Explanation:

Fundamental frequency,
$v = \frac{1}{2 l} \sqrt{\frac{T}{m}} = \frac{1}{2 l} \sqrt{\frac{T}{ρ r^{2}}} [∵ ρ = \frac{m}{r^{2}}]$
$\Rightarrow v \propto \frac{1}{l r}$
$∴ \frac{v_{1}}{v_{2}} = \frac{l_{2} r_{2}}{l_{1} r_{1}}$
$= \frac{2 L \times r}{L \times 2 r} = 1$

PHXI15:WAVES

355169 In a stationary wave

1 There is no net transfer of energy

2 Energy is constant at all points

3 Amplitude is same for all points

4 Energy and amplitude is same at all points

PHXI15:WAVES

355170 The equation of a stationary wave is $y = 2 \sin (\frac{π x}{15}) \cos (48 π t)$ . The distance between a node and its next antinode is

1 22.5 units

2 7.5 units

3 30 units

4 1.5 units

Explanation:

Comparing the given equation with standard equation of stationary wave
$y = A \sin (k x) \cos (ω t)$ ,
$k = \frac{2 π}{λ} = \frac{π}{15} \Rightarrow λ = 30$
Distance between a node and next antinode $= \frac{λ}{4} = \frac{30}{4} = 7.5$ units

KCET - 2019

PHXI15:WAVES

355171 A string vibrates with a frequency 200 $H z$ . Its length is doubled and its tension is altered until it begins to vibrate with frequency 300 $H z$ . What is the ratio of the new tension to the original tension?

1 3

2 9

3 12

4 5

Explanation:

Let $f_{1}$ and $f_{2}$ be the frequencies in two cases and $L_{1}, L_{2}$ and $T_{1}, T_{2}$ be respectively the corresponding lengths of the string and tensions under which it is vibrating.
$f_{1} = \frac{1}{2 L_{1}} \sqrt{\frac{T_{1}}{μ}}$
and $f_{2} = \frac{1}{2 L_{2}} \sqrt{\frac{T_{2}}{μ}}$
Thus, $\frac{f_{1}}{f_{2}} = \frac{L_{2}}{L_{1}} \sqrt{\frac{T_{1}}{T_{2}}}$ or $\frac{200}{300} = 2 \sqrt{\frac{T_{1}}{T_{2}}}$ $(f_{1} = 200, f_{2} = 300$ and $L_{2} = 2 L_{1}$ or $L_{2} / L_{1} = 2)$
or $\frac{2}{3} = 2 \sqrt{\frac{T_{1}}{T_{2}}}$ or $\frac{4}{9} = 4 \frac{T_{1}}{T_{2}}$ or $\frac{T_{2}}{T_{1}} = 9$

PHXI15:WAVES

355172 Two vibrating strings of the same material but lengths $L$ and $2 L$ have radii $2 r$ and $r$ respectively. They are stretched under the same tension. Both the strings vibrate in their fundamental modes, the one of length $L$ with frequency $v_{1}$ and the other with frequency $v_{2}$ . The ratio $v_{1} / v_{2}$ is

1 2

2 4

3 3

4 1

Explanation:

Fundamental frequency,
$v = \frac{1}{2 l} \sqrt{\frac{T}{m}} = \frac{1}{2 l} \sqrt{\frac{T}{ρ r^{2}}} [∵ ρ = \frac{m}{r^{2}}]$
$\Rightarrow v \propto \frac{1}{l r}$
$∴ \frac{v_{1}}{v_{2}} = \frac{l_{2} r_{2}}{l_{1} r_{1}}$
$= \frac{2 L \times r}{L \times 2 r} = 1$

PHXI15:WAVES

355169 In a stationary wave

1 There is no net transfer of energy

2 Energy is constant at all points

3 Amplitude is same for all points

4 Energy and amplitude is same at all points

PHXI15:WAVES

355170 The equation of a stationary wave is $y = 2 \sin (\frac{π x}{15}) \cos (48 π t)$ . The distance between a node and its next antinode is

1 22.5 units

2 7.5 units

3 30 units

4 1.5 units

Explanation:

Comparing the given equation with standard equation of stationary wave
$y = A \sin (k x) \cos (ω t)$ ,
$k = \frac{2 π}{λ} = \frac{π}{15} \Rightarrow λ = 30$
Distance between a node and next antinode $= \frac{λ}{4} = \frac{30}{4} = 7.5$ units

KCET - 2019

PHXI15:WAVES

355171 A string vibrates with a frequency 200 $H z$ . Its length is doubled and its tension is altered until it begins to vibrate with frequency 300 $H z$ . What is the ratio of the new tension to the original tension?

1 3

2 9

3 12

4 5

Explanation:

Let $f_{1}$ and $f_{2}$ be the frequencies in two cases and $L_{1}, L_{2}$ and $T_{1}, T_{2}$ be respectively the corresponding lengths of the string and tensions under which it is vibrating.
$f_{1} = \frac{1}{2 L_{1}} \sqrt{\frac{T_{1}}{μ}}$
and $f_{2} = \frac{1}{2 L_{2}} \sqrt{\frac{T_{2}}{μ}}$
Thus, $\frac{f_{1}}{f_{2}} = \frac{L_{2}}{L_{1}} \sqrt{\frac{T_{1}}{T_{2}}}$ or $\frac{200}{300} = 2 \sqrt{\frac{T_{1}}{T_{2}}}$ $(f_{1} = 200, f_{2} = 300$ and $L_{2} = 2 L_{1}$ or $L_{2} / L_{1} = 2)$
or $\frac{2}{3} = 2 \sqrt{\frac{T_{1}}{T_{2}}}$ or $\frac{4}{9} = 4 \frac{T_{1}}{T_{2}}$ or $\frac{T_{2}}{T_{1}} = 9$

PHXI15:WAVES

355172 Two vibrating strings of the same material but lengths $L$ and $2 L$ have radii $2 r$ and $r$ respectively. They are stretched under the same tension. Both the strings vibrate in their fundamental modes, the one of length $L$ with frequency $v_{1}$ and the other with frequency $v_{2}$ . The ratio $v_{1} / v_{2}$ is

1 2

2 4

3 3

4 1

Explanation:

Fundamental frequency,
$v = \frac{1}{2 l} \sqrt{\frac{T}{m}} = \frac{1}{2 l} \sqrt{\frac{T}{ρ r^{2}}} [∵ ρ = \frac{m}{r^{2}}]$
$\Rightarrow v \propto \frac{1}{l r}$
$∴ \frac{v_{1}}{v_{2}} = \frac{l_{2} r_{2}}{l_{1} r_{1}}$
$= \frac{2 L \times r}{L \times 2 r} = 1$

PHXI15:WAVES

355169 In a stationary wave

1 There is no net transfer of energy

2 Energy is constant at all points

3 Amplitude is same for all points

4 Energy and amplitude is same at all points

PHXI15:WAVES

355170 The equation of a stationary wave is $y = 2 \sin (\frac{π x}{15}) \cos (48 π t)$ . The distance between a node and its next antinode is

1 22.5 units

2 7.5 units

3 30 units

4 1.5 units

Explanation:

Comparing the given equation with standard equation of stationary wave
$y = A \sin (k x) \cos (ω t)$ ,
$k = \frac{2 π}{λ} = \frac{π}{15} \Rightarrow λ = 30$
Distance between a node and next antinode $= \frac{λ}{4} = \frac{30}{4} = 7.5$ units

KCET - 2019

PHXI15:WAVES

355171 A string vibrates with a frequency 200 $H z$ . Its length is doubled and its tension is altered until it begins to vibrate with frequency 300 $H z$ . What is the ratio of the new tension to the original tension?

1 3

2 9

3 12

4 5

Explanation:

Let $f_{1}$ and $f_{2}$ be the frequencies in two cases and $L_{1}, L_{2}$ and $T_{1}, T_{2}$ be respectively the corresponding lengths of the string and tensions under which it is vibrating.
$f_{1} = \frac{1}{2 L_{1}} \sqrt{\frac{T_{1}}{μ}}$
and $f_{2} = \frac{1}{2 L_{2}} \sqrt{\frac{T_{2}}{μ}}$
Thus, $\frac{f_{1}}{f_{2}} = \frac{L_{2}}{L_{1}} \sqrt{\frac{T_{1}}{T_{2}}}$ or $\frac{200}{300} = 2 \sqrt{\frac{T_{1}}{T_{2}}}$ $(f_{1} = 200, f_{2} = 300$ and $L_{2} = 2 L_{1}$ or $L_{2} / L_{1} = 2)$
or $\frac{2}{3} = 2 \sqrt{\frac{T_{1}}{T_{2}}}$ or $\frac{4}{9} = 4 \frac{T_{1}}{T_{2}}$ or $\frac{T_{2}}{T_{1}} = 9$

PHXI15:WAVES

355172 Two vibrating strings of the same material but lengths $L$ and $2 L$ have radii $2 r$ and $r$ respectively. They are stretched under the same tension. Both the strings vibrate in their fundamental modes, the one of length $L$ with frequency $v_{1}$ and the other with frequency $v_{2}$ . The ratio $v_{1} / v_{2}$ is

1 2

2 4

3 3

4 1

Explanation:

Fundamental frequency,
$v = \frac{1}{2 l} \sqrt{\frac{T}{m}} = \frac{1}{2 l} \sqrt{\frac{T}{ρ r^{2}}} [∵ ρ = \frac{m}{r^{2}}]$
$\Rightarrow v \propto \frac{1}{l r}$
$∴ \frac{v_{1}}{v_{2}} = \frac{l_{2} r_{2}}{l_{1} r_{1}}$
$= \frac{2 L \times r}{L \times 2 r} = 1$