Here,
y(x, t) = 5.0 sin (48t + 0.0264x + \(\pi\over 6\))
The general equation of a plane progressive wave is,
v(x,t) = a Sin\(\left[{2\pi\over \lambda}(vt+x)+\phi\right]\)
It is observed that the given equation represent a travelling waveform right to left.
Velocity \(V={48\over 0.0264}=1818.18cms^{-1}, r=5cm\)
(a) Amplitude and frequency: 
Amplitude,
\({2\pi\over \lambda}=0.0264\)
or
\(\lambda={2\pi\over 0.0264}cm={2\times3.14\over 0.0264}={6.28\over 0.0264}=237.8cm\)
frequency,
From the equation
v = ⋋v,
\(v={v\over \lambda}={1818.18\over 2\pi}\times0.0264\)
\(={1818.18\over 2\times3.14}\times0.0264\)
= 289.51 x 0.0264
= 7.64 Hz
b) To find least distance between two successive crests in the wave.
\(\lambda={2\pi\over 0.0264 }={2\times3.14\over 0.0264}={6.28\over0.0264}\)
= 237.8 em = 2.38 m
(c) When \(x={\lambda\over 4}\)
\(\phi={2\pi\over \lambda}\times{\lambda\over 4}={\pi\over 2}rad\)

Here,
y = 2.0 cos 2π(8t - 0.0060x + 0.27)
= 2.0 cos [2π (8t - 0.0060x) + 2π(0.27)]
Standard equation for a travelling wave is,
\(y=r\ cos\left[ {2\pi\over \lambda}(vt-x)+\phi\right]\)
Here
\(\phi={2\pi\over \lambda}x=2\pi\times0.006x\)
\({2\pi\over \lambda}=0.006\)
(a) When x = 300 em,
ψ= 2π\(\times\) 0.006\(\times\)300
=3.6π rad.
(b) When x = 0.75 m = 75 em,
ψ = 2π\(\times\)0.006\(\times\)75
= 0.9π rad
(c) When \(x={{\lambda}\over 4}\)
\(\phi={2\pi\over \lambda}\times{\lambda\over4}={\lambda \over 2}rad\)

Let total time t = t₁ + t₂, where t₁ is the time take from top to surface of water and t₂ is the time taken by sound to reach the top.
Calculation of t₁:
Using \(s=ut_1+{1\over 2}at_1^2\)
Putting, u = 0, a = g = 9.8 m/s²
S = 200 m , we get
\(200={1\over2}\times9.8\times t_1^2\)
200= 4.9x/2
t² = 40.816
t = 6.3887 s
and
Calculation of t₂:
Using \(v={s\over t_2}\)
\(t_2={s\over v}={200\over 3470}=0.588s\)
∴ Total time t = t₁ + t₂
= 6.3887 + 0.588
= 6.97 s
The splash has heard at 6.97s

Velocity of sound in air at 0°C = 331 mls.
Audible frequencies, V₁ = 20 Hz, V₂ = 20 x 10³ Hz.
(i) Period (T):
T·tme penod \(T={1\over V}\), we get
\(∵\ T_1={1\over V_1}={1\over 20}=0.05s\)
\(T_2={1\over V_2}={1\over 20\times10^3}=0.055\times 10^{-3}s\)
∴ Time period range is 0.05 to 0.055 \(\times\)10^-3 s.
(ii) Wavelength ⋋:
From the formula \(\lambda ={v\over \lambda}\)
∴ \(\lambda_1={v\over V_1}={331\over 20}=16.55m\)
and
\(\lambda _2={v\over V_2}={331\over 20\times10^3}=0.0165m\)
∴ Wave length range is 16.55 m to 0.0165 m
(iii) Angular frequency ω:
From the formula, 0 = 27tV
∴ ω = 2πV₁= 2π \(\times\)20
= 40π rads^-1
and ω₂ = 2πV₂= 2π\(\times\)20\(\times\)10³
= 40π\(\times\)10³ rad/s.
∴  Angular frequency range is 40π to 40π\(\times\)10³ rad/s

(a) Stethoscope: It works on the, principle of multiple reflections.

It consists of three main parts:
(i) Chest piece
(ii) Ear piece
(iii) Rubber tube
(i) Chest piece: It consists of a small disc-shaped resonator (diaphragm) which is very sensitive to sound and amplifies the sound it detects.
(ii) Ear piece: It is rnade up of metal tubes which are used to hear sounds detected by the chest piece.
(iii) Rubber tube: This tube connects both chest piece and ear piece. It is used to transmit the sound signal detected by the diaphragm, to the ear piece. The sound of heart beats (or lungs) or any sound produced by internal organs can be detected, and it reaches the ear piece through this tube by multiple reflections.
(b) Echo: An echo is a repetition of sound produced by the reflection of sound waves from a wall, mountain or other obstructing surfaces. The speed of sound in air at 20°C is 344 ms^-1 If we shout at a wall which is, at 344 m away, then the sound will take 1 second to reach the wall. After reflection, the sound will take one more second to reach us. Therefore, we hear the echo after, two seconds.
Scientists have estimated that we can hear two sounds properly if the time gap or time \(\left(1\over 10\right)^{th}\) of a second (persistence of hearing) i.e., 0.1 s
Then, 
\(velocity={Distance\ travelled\over time\ taken}=2d\)
2d= 344\(\times\)0.1 = 34.4 m
d= 17.2 m.
The minimum distance from a sound reflecting wall to hear an echo at 20°C is 17.2 meter.
(c) SONAR: Sound Navigation and Ranging. Sonar systems make use of reflections of sound waves in water to locate the position or motion of an object. Similarly, dolphins and bats use the sonar principle to find their way in the darkness
(d) Reverberation: In a closed room the sound is repeatedly reflected from the walls and it is even heard long after the sound source ceases to function. The residual sound remaining in an enclosure and the phenomenon of multiple reflections of sound is called reverberation. The duration for which the sound persists is called reverberation time. It should be noted that the reverberation time greatly affects the quality of sound heard in a hall. Therefore, halls are constructed with some optimum reverberation time.