The distance between successive nodes or anti-nodes is \(\frac{\lambda}{2}\).The wavelength (λ) of the light is of the order of 10^-7m, so the distance between successive nodes or antinodes is also of the order of 10^-7 m. Since this distance is very small and cannot be detected by the eye or by an ordinary optical instrument, Hence nodes and antinodes are not seen.

No, the converse is not true. The basic requirement for a wave function to represent a travelling wave is that for all values of x and t, wave function must have a finite value. Out of the given functions for y none satisfies this condition. Therefore, none can represent a travelling wave.

Give, v = 1000 kHz =10⁶ Hz
v_a = 340 m/s, v_w = 1486 m/s
Wavelength of reflected sound, \({ \lambda }_{ a }=v\frac { { v }_{ a } }{ v } \)
\(=\frac { 340 }{ { 10 }^{ 6 } } =3.4\times { 10 }^{ -4 }m\)      
Wavelength of transmitted sound,
\({ \lambda }_{ w }=\frac { { v }_{ s } }{ v } =\frac { 1486 }{ { 10 }^{ 6 } } =1486\times { 10 }^{ -6 }\)     
\({ \lambda }_{ w }=1.486\times { 10 }^{ -3 }m\)

On reflection from the denser medium, there will be a phase change of 180⁰
Net amplitude = \(\frac { 2 }{ 3 } \times 0.6=0.4\)
Hence, equation of reflected wave will be 
y = 0.4sin\(2\pi \left[ t+\frac { x }{ 2 } +\pi \right] \)
\( =0.4sin2\pi (t+{ x }/{ 2 })\)

The sonometer frequency is given by
v=\(\frac{n}{2L}\sqrt{\frac{T}{μ}}\)
Now, as it vibrates with length L, we assume v=v₁
n = n₁
∴v₁=\(\frac{n_1}{2L}\sqrt{\frac{T}{μ}}\)
When length is doubled, then
v₂=\(\frac{n_2}{2 \times2L}\sqrt{\frac{T}{μ}}\)
Dividing Eq.(i) by Eq. (ii), we get
\(\frac{v_1}{v_2}=\frac{n_1}{n_2}×2\)
To keep the resonance,
\(\frac{v_1}{v_2}=1=\frac{n_1}{n_2}×2\)
⇒n₂=2n₁
Hence, when the wire is doubled, the number of loops also get doubled to produce the resonance. That is, it resonates in second harmonic.

In case of a string fixed at two ends, when the string vibrates in n loops
v_n= \(\frac{n}{2l}\sqrt{\frac{T}{μ}}\) ⇒v_n ∝ n
Hence, when the string vibrates in 1 loop, 2 loops, 3 loops, 4 loops, the frequencies are in the ratio 1:2:3:4.

Water waves produced by a motorboat sailing in water are both longitudinal and transverse.

1031.25 Hz

At t = 0, \(y=\frac{1}{1+x^{2}}\)
∴ \(1+x^{2}=\frac{1}{y}\)
\(x^{2}=\frac{1}{y}-1=\frac{1-y}{y} x= (\frac{1-y}{y})^{\frac{1}{2}}\)
At t = 2s, \(y=\frac{1}{[1+(x+1)^{2}]}\)
\(1+(x-1)^{2} =\frac{1}{y}\)
\((x-1)^{2} = \frac{1}{y}-1 = \frac{1-y}{y}\)
\((x-1)=(\frac{1-y}{y})^{\frac{1}{2}}\)
⇒ \(x=1+(\frac{1-y}{y})^{\frac{1}{2}}\)
Since, \(v=\frac{x_2-x_1}{t_2-t_1}\)
∴ \(v=\frac{1}{2-0}=0.5 ms^{-1}\)

Here, given v = 400Hz
\({ v }_{ w }=10m/s\) = speed of wind 
speed of sound in still air = 340 m/s
As the wind is blowing in the same direction as wave hence, effective speed of sound
\(=v+{ v }_{ w }=340+10=350m/s\) On the platform as both source and observer are at rest, hence, frequency remains unchanged v = 400 Hz.
Wavelength,\(\lambda =\frac { v+{ v }_{ w } }{ v } \frac { 350 }{ 400 } =0.875\quad m\)
When air is still \({ v }_{ w }=0\) observer's speed = \({ v }_{ 0 }\) = 10\({ ms }^{ -1 }\);\({ v }_{ s }=0\)As observer moves towards the source
\({ v }^{ ' }=\frac { v+{ v }_{ 0 } }{ v } \times v\)
\( \Rightarrow { v }^{ ' }\left( \frac { 340+10 }{ 340 } \right) \times 400\)
\( =\left( \frac { 350 }{ 340 } \right) \times 400\)
\(\\ =\left( \frac { 35 }{ 34 } \right) (400)=411.76\ Hz\)
As source is at rest wavelength dos not change, 
Speed of sound will remain same = 340m/s
It is obvious that both the cases are different.

Given, y(x,t) = 0.005 sin (80.0x - 3.0t)
Then, y(x,t) = asin (kx-\(\omega \)t)
Now, compare the given equation with standard equation to find out all the physical quantities.
a = 0.005 m
k = 80.0 rad/m
 \(\omega \) = 3.0 rad/s
The physical quantities by using the given fundamental physical quantities
(ii) Wavelength, \(\lambda =\frac { 2\pi }{ k } =\frac { 2\pi }{ 80 } =7.85\quad cm\)
Now, the displacement y of the particle at a distance.
x = 30.0cm = 0.3 m snd time t =20 s
y(0.3,20) = 0.005 sin (80\(\times \)0.3-3.0\(\times \)20)
y(0.3,20) = 0.005 sin (24 -60)
= 0.00495 m\(\simeq \) 5mm

(a) Transverse and longitudinal
(b) Longitudinal
(c) Transverse and longitudinal
(d) Longitudinal

(i) Values displayed are: Creative mind, awareness, scientific interest and keen observer.
(ii) Here v = 350 m/s, l = 0.5 m
∴ For closed pipe fundamental frequency
\(v=\frac{v}{4l}=\frac{350}{4\times 0.5}=175 Hz\)
Frequency of second overtone = 5v=\(5 \times 175 = 875 Hz\)

(a) Here nv = 280 Hz
and (n + 1)v = 350 Hz
∴ (n + 1) v - nv 350 - 280 = 70
v = 70 Hz
(b) Here \(\frac{a_2}{a_1}\)=3
∴ \(\frac{I_2}{I_1}=(\frac{a_2}{a_1})^{2}=9\)
Now, log₁₀ \((\frac{I_2}{I-1})=10 log_{10 }(\frac{I_2}{I_1})\)
= 10 log₁₀(9)
= 10 log₁₀ (3)² = 20 log₁₀ ³
= log₁₀ 3²⁰  
ஃ \(\frac{I_2}{I_1}=3^{20}\ or \ I_2 = 3^{20} I_1.\)

Given L = 20 cm = 0.2 m, v_n = 430 Hz, v = 340 m/s
It will behave as closed organ pipe
\( { v }_{ n }=\left( 2n-1 \right) \frac { v }{ 4L } ,\ where\ n=1,2,3....\)
\( \Rightarrow 430=\left( 2n-1 \right) \frac { v }{ 4L } =\left( 2n-1 \right) \times \frac { 340 }{ 4\times 0.2 } \)
\(\Rightarrow \ \left( 2n-1 \right) =\frac { 430\times \left( 0.8 \right) }{ 340 } \Rightarrow 2n=\frac { \left( 430 \right) \left( 0.8 \right) }{ 340 } +1\)
\( \Rightarrow n=\frac { 43\times 4 }{ 340 } +\frac { 1 }{ 2 } =\frac { 2\times 172+340 }{ 340\times 2 } =\frac { 684 }{ 680 } =1.006\)
Hence, it will be the 1st normal mode or harmonic mode of vibration.
In a pipe open at both ends,\({ v }_{ n }=n\times \frac { v }{ 2l } =\frac { n\times 340 }{ 2\times 0.2 } =430\) 
\(\Rightarrow n=\frac { 430\times 2\times 0.2 }{ 340 } =\frac { 43\times 2\times 2 }{ 340 } =0.5\)
As n is not integer, hence, open organ pipe cannot be in resonance with the source.

Given, h = 300m, g = 9.8m/s^2 , v = 340ms^-1
t₁ = time taken by stone to strike the water surface 
 \(t_{ 1 }=\sqrt { \frac { 2h }{ g } } =\sqrt { \frac { 300 }{ 49 } } =7.82s\left( as\quad h=0+\frac { 1 }{ 2 } gt^{ 2 }_{ 1 } \right) \)
t₂ = time taken by the splash's sound to reach top of the tower
\(t_{ 2 }=\frac { h }{ v } =\frac { 300 }{ 340 } =0.882\quad \left[ v=\frac { h }{ t_{ 2 } } \right] \)
Total time, t = time to hear splash of sound 
= t₁ + t₂ =7.82 + 0.882
= 8.702