Explanation​:
Since it is pulled down by 10 $cm$, that is the maximum displacement and hence the amplitude.

If it takes 0.5 s to return to the equilibrium position, then the complete oscillation is $=2s$.

Since f=T1​ and $T=2s$, frequency is 0.5 Hz

Explanation​:
Waves are produced in the same medium. This means they have the same speed.

They also have the same period, implying that their wavelengths are also equal.

Since the waves start half a period apart, the phase difference between the sources is half a cycle, and remains constant since they are produced at the same frequency.

Point $C$ is five wavelengths from $A$, and four wavelengths from $B$, making the path difference one wavelength or a whole cycle. But after every single cycle, the two waves maintain a constant phase difference of $\pi$, leading to destructive interference.

Explanation​:
The principle of superposition states that the resultant wave is the sum of displacement from equilibrium of each wave.

The amplitude of the wave is the maximum displacement from equilibrium, which can be read off the graph as $7cm-5cm=2cm$.

Since each of the waves have an amplitude of 2cm and the equilibrium water level is $5cm$, after superimposing the two waves we find that the amplitude of the resultant wave is $4cm$.

Therefore, the maximum and minimum water levels can be determined as: 5cm+4 cm=9 cm for the maximum level and $5cm-4cm=1cm$ for the minimum level.

Here, the source of the travelling wave is the vibration of the speaker up and down, and the medium of that wave is the string.

Statement I is correct. A decrease in the frequency of the sound from the speaker results in a decrease in the frequency of the travelling wave in the string. The wave speed is constant, as the medium of the travelling wave remains the string. The wavelength is inversely proportional to the frequency according to the wave equation:

v=λf→λ=v/f​

As a result, the wavelength must increase.

Statement II is also correct. A louder sound means that the amplitude of the longitudinal sound wave in the air is larger. This is caused by greater amplitude vibrations from the speaker, which also produces larger amplitude vibrations in the string.

Statement III is incorrect. As previously mentioned, the wave speed is constant as the medium remains the same. Even though the frequency of the wave has increased, the wavelength will decrease in proportion to keep the wave speed equal to the wavelength times the frequency.

Therefore, the correct answer is B, I and II only.

Explanation​:
Recall for simple harmonic motion, a∝−x, therefore the object experiences maximum negative acceleration at maximum positive displacement from equilibrium.

Explanation​:
Two conditions must be satisfied to define motion as simple harmonic motion;

• The acceleration is proportional to the displacement.
• The acceleration and displacement are in opposite directions.

The second condition indicates the existence of a negative sign in front of the displacement and the first condition demands to have $x$ instead of x2 in the expression. These two conditions are satisfied in B.

Explanation​:

Doppler effect only occurs when there is relative motion between the observer and the source. In this case, the satellite and receiver are motionless relative to each other. Therefore, there will be no change in frequency.

Explanation​:
The wave is faster in medium $Y$ than in medium $X$. As a result, the ray representing the wave is expected to bend away from the normal, refracting into the original medium.

If the angle of incidence exceeds the critical angle, however, the angle of refraction will be undefined and the wave will undergo total internal reflection.

This is exactly what is shown by Option D, which is the correct response.

Option A shows the ray bending towards the normal, Option C shows no bending at all, and Option B shows an impossible path for the wave to travel.

Explanation​:
Light will refract away from the normal at a boundary where there is an increase in speed as in a glass to air boundary. Answers A and C show this. A light ray will not refract at a boundary if its angle of incidence is 0°. It must therefore enter the curved edge aligned with the center of the straight side as in A. Therefore the answer is A.

Explanation​:
Diffraction is the spreading out of a wave while passing through an obstacle or aperture. The door of the room can be considered as the aperture. When the wavelength of the wave is comparable with the aperture size, diffraction is greater. The wavelength of the sound wave is much larger than the wavelength of the light wave and closer to the size of the door. Therefore, sound waves diffract more while passing through the door and reach the student while light waves do not. The answer is A.

Explanation​:

The distance between wavefronts is equal to the wavelength. Since the wavefronts towards the child are more separated than wavefronts in the other direction, the child will receive wavefronts less frequently. Therefore, the child hears the sound with less frequency than $f$. The observed frequency is less than the frequency of the wave created by the source when the source and observer are receding from each other.

The amplitude is $4cm$ on either side of the equilibrium position because in one full cycle it travels four amplitudes (16/4=4 cm​).

Explanation​:
A wave travelling on a rope is an example of a transverse wave, where particles of the medium move up and down in a direction perpendicular to the direction of propagation. As the wave moves to the right, particle $P$ oscillates only up and down.

After half a period, the particle will be in the position shown below.

The wave is moving to the right. This means that a very short time after the current position, the solid line representing the waveform will be higher at the current position of $P$. As a result, point $P$ is moving upwards, and the velocity vector will be directed upwards.

To determine the magnitude of the velocity, we can recognize that because particle P has moved through half a period of its cycle, particle $P$ will be displaced from equilibrium the exact same distance as before - half of the amplitude.

This means that the particle will be moving at the exact same speed as before.

Explanation​:
For a circle of diameter $d$, circumference is $\pi d$. Thus the cicrumference in this case is $15\pi$. In this instance the distance travelled by the bob is exactly equal to twice its circumference. The shadow, therefore, has travelled two complete oscillations and is at exactly the same point at which it started. The displacement, therefore, is zero.