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PHXI15:WAVES

355152 A wire under tension vibrates with a fundamental frequency of $600 H z$ . If the length of the wire is doubled, the radius is halved and the wire is made to vibrate under one-ninth the tension. Then the fundamental frequency will become

1

200 H z

2

300 H z

3

600 H z

4

400 H z

Explanation:

The fundamental frequency of vibration is
$f = \frac{1}{2 L r} \sqrt{\frac{T}{π ρ}}$
Where $T =$ Tension in the wire
$L =$ Length of the wire
$r =$ Radius of the wire
$ρ =$ Density of material of the wire
$\begin{aligned} ∴ \frac{f_{1}}{f_{2}} = (\frac{L_{2}}{L_{1}}) (\frac{r_{2}}{r_{1}}) \sqrt{\frac{T_{1}}{T_{2}}} \\ ∴ \frac{f_{1}}{f_{2}} = (\frac{2 L_{1}}{L_{1}}) (\frac{\frac{1}{2} r_{1}}{r_{1}}) \sqrt{\frac{T_{1}}{T_{1} / 9}} = 2 \times \frac{1}{2} \times 3 = 3 \end{aligned}$
$\Rightarrow f_{2} = \frac{f_{1}}{3} = \frac{600 H z}{3} = 200 H z$

KCET - 2011

PHXI15:WAVES

355153 The tension in a stretched string fixed at both ends is changed by $2 %$ , the fundamental frequency is found to get changed by $15 H z$ . Select the incorrect statement.

1 Wavelength of the string of fundamental frequency does not change

2 Velocity of propagation of wave changes by

2 %

3 Velocity of propagation of wave changes by

1 %

4 Original frequency is

1500 H z

Explanation:

$f_{0} =$ fundamental frequency $= \frac{1}{2 l} \sqrt{\frac{T}{μ}}$
$\frac{Δ f_{0}}{f_{0}} = \frac{1}{2} \cdot \frac{Δ T}{T} \Rightarrow f_{0} = \frac{2 \times Δ f_{0}}{(Δ T / T)} = \frac{2 \times 15}{(2 / 100)}$
$= 1500 H z$
Also, $v = \sqrt{\frac{T}{μ}} ∴ \frac{Δ v}{v} = \frac{1}{2} \cdot \frac{Δ T}{T} = \frac{1}{2} \times 2 % = 1 %$
For fundamental frequency,
$\frac{λ}{2} = l \Rightarrow λ = 2 l = constant .$
So option (2) is correct.

PHXI15:WAVES

355154 In a standing wave experiment, a $1.2 k g$ horizontal rope is fixed in place at its two ends( $x$ $= 0$ and $x = 2.0 m$ ) and made to oscillate up and down in the fundamental mode, at a frequency of $5.0 H z$ . At $t = 0$ , the point at $x = 1.0 m$ has zero displacement and is moving upward in the positive direction of $y$ - axis with a transverse velocity $3.14 m / s$ . The standing wave equation is

1

y = 0.1 \sin (\frac{π}{2} x) \sin (10 π t)

2

y = 0.1 \sin (π x) \sin (10 π t)

3

y = 0.05 \sin (\frac{π}{2} x) \cos (10 π t)

4

y = 0.04 \sin (\frac{π}{2} x) \sin (10 π t)

Explanation:

The wave is in fundamental mode of vibration as shown in the figure
The angular frequency
$ω = 2 π v = 10 π$
supporting img

$\frac{λ}{2} = l \Rightarrow k = \frac{2 π}{λ} = \frac{π}{l}$
The velocity of the particle at the mid point of the string is $v_{p} = A ω = 3.14 m / s \Rightarrow A = 0.1 m$
From the above information
$y = 0.1 \sin (\frac{π}{2} x) \sin (10 π t)$

PHXI15:WAVES

355155 In a stationary wave

1 Amplitude is zero at antinodes

2 Amplitude is zero at nodes

3 Amplitude is maximum at nodes

4 Amplitude is zero at all points.

PHXI15:WAVES

355152 A wire under tension vibrates with a fundamental frequency of $600 H z$ . If the length of the wire is doubled, the radius is halved and the wire is made to vibrate under one-ninth the tension. Then the fundamental frequency will become

1

200 H z

2

300 H z

3

600 H z

4

400 H z

Explanation:

The fundamental frequency of vibration is
$f = \frac{1}{2 L r} \sqrt{\frac{T}{π ρ}}$
Where $T =$ Tension in the wire
$L =$ Length of the wire
$r =$ Radius of the wire
$ρ =$ Density of material of the wire
$\begin{aligned} ∴ \frac{f_{1}}{f_{2}} = (\frac{L_{2}}{L_{1}}) (\frac{r_{2}}{r_{1}}) \sqrt{\frac{T_{1}}{T_{2}}} \\ ∴ \frac{f_{1}}{f_{2}} = (\frac{2 L_{1}}{L_{1}}) (\frac{\frac{1}{2} r_{1}}{r_{1}}) \sqrt{\frac{T_{1}}{T_{1} / 9}} = 2 \times \frac{1}{2} \times 3 = 3 \end{aligned}$
$\Rightarrow f_{2} = \frac{f_{1}}{3} = \frac{600 H z}{3} = 200 H z$

KCET - 2011

PHXI15:WAVES

355153 The tension in a stretched string fixed at both ends is changed by $2 %$ , the fundamental frequency is found to get changed by $15 H z$ . Select the incorrect statement.

1 Wavelength of the string of fundamental frequency does not change

2 Velocity of propagation of wave changes by

2 %

3 Velocity of propagation of wave changes by

1 %

4 Original frequency is

1500 H z

Explanation:

$f_{0} =$ fundamental frequency $= \frac{1}{2 l} \sqrt{\frac{T}{μ}}$
$\frac{Δ f_{0}}{f_{0}} = \frac{1}{2} \cdot \frac{Δ T}{T} \Rightarrow f_{0} = \frac{2 \times Δ f_{0}}{(Δ T / T)} = \frac{2 \times 15}{(2 / 100)}$
$= 1500 H z$
Also, $v = \sqrt{\frac{T}{μ}} ∴ \frac{Δ v}{v} = \frac{1}{2} \cdot \frac{Δ T}{T} = \frac{1}{2} \times 2 % = 1 %$
For fundamental frequency,
$\frac{λ}{2} = l \Rightarrow λ = 2 l = constant .$
So option (2) is correct.

PHXI15:WAVES

355154 In a standing wave experiment, a $1.2 k g$ horizontal rope is fixed in place at its two ends( $x$ $= 0$ and $x = 2.0 m$ ) and made to oscillate up and down in the fundamental mode, at a frequency of $5.0 H z$ . At $t = 0$ , the point at $x = 1.0 m$ has zero displacement and is moving upward in the positive direction of $y$ - axis with a transverse velocity $3.14 m / s$ . The standing wave equation is

1

y = 0.1 \sin (\frac{π}{2} x) \sin (10 π t)

2

y = 0.1 \sin (π x) \sin (10 π t)

3

y = 0.05 \sin (\frac{π}{2} x) \cos (10 π t)

4

y = 0.04 \sin (\frac{π}{2} x) \sin (10 π t)

Explanation:

The wave is in fundamental mode of vibration as shown in the figure
The angular frequency
$ω = 2 π v = 10 π$
supporting img

$\frac{λ}{2} = l \Rightarrow k = \frac{2 π}{λ} = \frac{π}{l}$
The velocity of the particle at the mid point of the string is $v_{p} = A ω = 3.14 m / s \Rightarrow A = 0.1 m$
From the above information
$y = 0.1 \sin (\frac{π}{2} x) \sin (10 π t)$

PHXI15:WAVES

355155 In a stationary wave

1 Amplitude is zero at antinodes

2 Amplitude is zero at nodes

3 Amplitude is maximum at nodes

4 Amplitude is zero at all points.

PHXI15:WAVES

355152 A wire under tension vibrates with a fundamental frequency of $600 H z$ . If the length of the wire is doubled, the radius is halved and the wire is made to vibrate under one-ninth the tension. Then the fundamental frequency will become

1

200 H z

2

300 H z

3

600 H z

4

400 H z

Explanation:

The fundamental frequency of vibration is
$f = \frac{1}{2 L r} \sqrt{\frac{T}{π ρ}}$
Where $T =$ Tension in the wire
$L =$ Length of the wire
$r =$ Radius of the wire
$ρ =$ Density of material of the wire
$\begin{aligned} ∴ \frac{f_{1}}{f_{2}} = (\frac{L_{2}}{L_{1}}) (\frac{r_{2}}{r_{1}}) \sqrt{\frac{T_{1}}{T_{2}}} \\ ∴ \frac{f_{1}}{f_{2}} = (\frac{2 L_{1}}{L_{1}}) (\frac{\frac{1}{2} r_{1}}{r_{1}}) \sqrt{\frac{T_{1}}{T_{1} / 9}} = 2 \times \frac{1}{2} \times 3 = 3 \end{aligned}$
$\Rightarrow f_{2} = \frac{f_{1}}{3} = \frac{600 H z}{3} = 200 H z$

KCET - 2011

PHXI15:WAVES

355153 The tension in a stretched string fixed at both ends is changed by $2 %$ , the fundamental frequency is found to get changed by $15 H z$ . Select the incorrect statement.

1 Wavelength of the string of fundamental frequency does not change

2 Velocity of propagation of wave changes by

2 %

3 Velocity of propagation of wave changes by

1 %

4 Original frequency is

1500 H z

Explanation:

$f_{0} =$ fundamental frequency $= \frac{1}{2 l} \sqrt{\frac{T}{μ}}$
$\frac{Δ f_{0}}{f_{0}} = \frac{1}{2} \cdot \frac{Δ T}{T} \Rightarrow f_{0} = \frac{2 \times Δ f_{0}}{(Δ T / T)} = \frac{2 \times 15}{(2 / 100)}$
$= 1500 H z$
Also, $v = \sqrt{\frac{T}{μ}} ∴ \frac{Δ v}{v} = \frac{1}{2} \cdot \frac{Δ T}{T} = \frac{1}{2} \times 2 % = 1 %$
For fundamental frequency,
$\frac{λ}{2} = l \Rightarrow λ = 2 l = constant .$
So option (2) is correct.

PHXI15:WAVES

355154 In a standing wave experiment, a $1.2 k g$ horizontal rope is fixed in place at its two ends( $x$ $= 0$ and $x = 2.0 m$ ) and made to oscillate up and down in the fundamental mode, at a frequency of $5.0 H z$ . At $t = 0$ , the point at $x = 1.0 m$ has zero displacement and is moving upward in the positive direction of $y$ - axis with a transverse velocity $3.14 m / s$ . The standing wave equation is

1

y = 0.1 \sin (\frac{π}{2} x) \sin (10 π t)

2

y = 0.1 \sin (π x) \sin (10 π t)

3

y = 0.05 \sin (\frac{π}{2} x) \cos (10 π t)

4

y = 0.04 \sin (\frac{π}{2} x) \sin (10 π t)

Explanation:

The wave is in fundamental mode of vibration as shown in the figure
The angular frequency
$ω = 2 π v = 10 π$
supporting img

$\frac{λ}{2} = l \Rightarrow k = \frac{2 π}{λ} = \frac{π}{l}$
The velocity of the particle at the mid point of the string is $v_{p} = A ω = 3.14 m / s \Rightarrow A = 0.1 m$
From the above information
$y = 0.1 \sin (\frac{π}{2} x) \sin (10 π t)$

PHXI15:WAVES

355155 In a stationary wave

1 Amplitude is zero at antinodes

2 Amplitude is zero at nodes

3 Amplitude is maximum at nodes

4 Amplitude is zero at all points.

PHXI15:WAVES

355152 A wire under tension vibrates with a fundamental frequency of $600 H z$ . If the length of the wire is doubled, the radius is halved and the wire is made to vibrate under one-ninth the tension. Then the fundamental frequency will become

1

200 H z

2

300 H z

3

600 H z

4

400 H z

Explanation:

The fundamental frequency of vibration is
$f = \frac{1}{2 L r} \sqrt{\frac{T}{π ρ}}$
Where $T =$ Tension in the wire
$L =$ Length of the wire
$r =$ Radius of the wire
$ρ =$ Density of material of the wire
$\begin{aligned} ∴ \frac{f_{1}}{f_{2}} = (\frac{L_{2}}{L_{1}}) (\frac{r_{2}}{r_{1}}) \sqrt{\frac{T_{1}}{T_{2}}} \\ ∴ \frac{f_{1}}{f_{2}} = (\frac{2 L_{1}}{L_{1}}) (\frac{\frac{1}{2} r_{1}}{r_{1}}) \sqrt{\frac{T_{1}}{T_{1} / 9}} = 2 \times \frac{1}{2} \times 3 = 3 \end{aligned}$
$\Rightarrow f_{2} = \frac{f_{1}}{3} = \frac{600 H z}{3} = 200 H z$

KCET - 2011

PHXI15:WAVES

355153 The tension in a stretched string fixed at both ends is changed by $2 %$ , the fundamental frequency is found to get changed by $15 H z$ . Select the incorrect statement.

1 Wavelength of the string of fundamental frequency does not change

2 Velocity of propagation of wave changes by

2 %

3 Velocity of propagation of wave changes by

1 %

4 Original frequency is

1500 H z

Explanation:

$f_{0} =$ fundamental frequency $= \frac{1}{2 l} \sqrt{\frac{T}{μ}}$
$\frac{Δ f_{0}}{f_{0}} = \frac{1}{2} \cdot \frac{Δ T}{T} \Rightarrow f_{0} = \frac{2 \times Δ f_{0}}{(Δ T / T)} = \frac{2 \times 15}{(2 / 100)}$
$= 1500 H z$
Also, $v = \sqrt{\frac{T}{μ}} ∴ \frac{Δ v}{v} = \frac{1}{2} \cdot \frac{Δ T}{T} = \frac{1}{2} \times 2 % = 1 %$
For fundamental frequency,
$\frac{λ}{2} = l \Rightarrow λ = 2 l = constant .$
So option (2) is correct.

PHXI15:WAVES

355154 In a standing wave experiment, a $1.2 k g$ horizontal rope is fixed in place at its two ends( $x$ $= 0$ and $x = 2.0 m$ ) and made to oscillate up and down in the fundamental mode, at a frequency of $5.0 H z$ . At $t = 0$ , the point at $x = 1.0 m$ has zero displacement and is moving upward in the positive direction of $y$ - axis with a transverse velocity $3.14 m / s$ . The standing wave equation is

1

y = 0.1 \sin (\frac{π}{2} x) \sin (10 π t)

2

y = 0.1 \sin (π x) \sin (10 π t)

3

y = 0.05 \sin (\frac{π}{2} x) \cos (10 π t)

4

y = 0.04 \sin (\frac{π}{2} x) \sin (10 π t)

Explanation:

The wave is in fundamental mode of vibration as shown in the figure
The angular frequency
$ω = 2 π v = 10 π$
supporting img

$\frac{λ}{2} = l \Rightarrow k = \frac{2 π}{λ} = \frac{π}{l}$
The velocity of the particle at the mid point of the string is $v_{p} = A ω = 3.14 m / s \Rightarrow A = 0.1 m$
From the above information
$y = 0.1 \sin (\frac{π}{2} x) \sin (10 π t)$

PHXI15:WAVES

355155 In a stationary wave

1 Amplitude is zero at antinodes

2 Amplitude is zero at nodes

3 Amplitude is maximum at nodes

4 Amplitude is zero at all points.

Explanation:

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