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.

Node: It is a point on stationary wave at which amplitude of vibration of the particle is zero.
Antinode: It is a point on stationary wave at which amplitude of vibration of the particle is maximum.

The beats cannot be heard due to persistence of hearing if the difference in frequencies is more than 10.

Velocity of sound in air increases by 0.61 m/ s, when temperature rises by 1⁰C.

This is because wood has high damping.

The time during which the intensity of sound decreases to 10^-6 times its original intensity.

As \(v \propto \sqrt{T}\) , therefore, wave speed becomes twice.

This is due to the difference in the frequency of the incident wave and the apparent frequency of the reflected wave.

Speed of transverse waves setup in a string depends upon the tension (T) in the string and the linear mass density (μ) of the string. In fact, \(v= \sqrt{\frac{T}{\mu}}\).

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.

We know that, speed,  \(v\infty \sqrt { T } \)
By formula v = \(\frac { xRT }{ p } \)
Where T is in kelvin 
\(\frac { v_{ t } }{ v_{ 0 } } =\sqrt { \frac { 273+t }{ 273+0 } } =3\)
\(\\ \Rightarrow \frac { 273+t }{ 273 } =9\Rightarrow \ t=9\times 273-273=2184^{ 0 }C\)

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\(2\pi \left[ t+\frac { x }{ 2 } +\pi \right] \)
\( =0.4sin2\pi (t+{ x }/{ 2 })\)

Length of pipe(l) = 20 cm = 20 x 10^-2 m
\({ v }_{ funda }=\frac { v }{ 4L } =\frac { 330 }{ 4\times 20\times { 10 }^{ -2 } }\)
\( \\ { v }_{ funda }=\frac { 330\times 100 }{ 80 } =412.5\quad Hz\)
\(\\ \frac { { v }_{ given } }{ { v }_{ funda } } =\frac { 1237.5 }{ 412.5 } =3\)
Hence, 3rd harmonic mode of the pipe is resonantly excited by the source of given frequency.

l = 12m, M = 2.10 kg, T = 2.06\(\times \)10⁴N, v =?
\(\mu =\frac { M }{ l } =\frac { 2.10 }{ 12 } kg/m\)
\(\\ v=\sqrt { \frac { T }{ \mu } } =\sqrt { \frac { 2.06\times { 10 }^{ 4 } }{ 2.10/12 } } =3.43\times { 10 }^{ 2 }m/s\)

1031.25 Hz