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 $(0.5)(4)=2s$.

Since f=1/T and $T=2s$, frequency is 0.5Hz.

Explanation​:

• f′=f(v/[v−us])

  Where f′= frequency detected by observer

  $f$ = source frequency

  $v$ = speed of sound in still air

The observed frequency is higher than the emitted frequency as the wavelength shortens due to the motion of the train. As the train speed does not change, the observed frequency will remain constant, although its intensity will increase.

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.

Explanation​:
It can be seen on the diagram that when the waves cross the boundary, the wavelength decreases. At a boundary, the frequency of a wave is unchanged because the number of waves entering the boundary every second must equal the number of waves that leave the boundary every second.

From the relationship v=fλ, if f remains constant and $\lambda$ decreases, the $v$ must also decrease.

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​:
Sound travels faster in water. According to Snell's law, the waves will refract away from the normal as seen in options $A$ and $C$.

Because $v=f\lambda$ and f is constant across the boundary, favelength also increases when the speed increases as seen in choices $B$ and C.

Therefore C is the correct answer.

The interference pattern is not modulated by the single slit diffraction when slit width is negligible. Therefore, the intensities for all maxima should be the same. A decrease in slit width results in a decrease of maximum intensity. Therefore the peak values should be smaller.

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

Explanation​:

With the increase of slits, more light passes through the diffraction grating and incident on the screen. Therefore, the intensity of light increases. The width of the primary maxima is decreased with an increase in the number of slits.

Explanation​:
The y axis cannot represent the velocity of the pendulum, as a velocity-position graph would also show negative velocities to represent motion in both directions.

The formula for kinetic energy is given by 1/2mω^2(0x^2 −x2), which would be parabolic in shape, opening downwards as is shown in the graph.

The potential energy of the bob should be less at the equilibrium position than in the positions where displacement is maximum.

Explanation​:
Speed is always positive and it should have maximum and minimum values twice in one complete oscillation. Therefore the answer is C.

Explanation​:
Acceleration is proportional to the resultant force. Therefore, force and acceleration must be in the same direction.

Explanation​:
To calculate the wave speed, we can use the wave equation:

v=λf

Since

• $\to$ v=λ/T​

The wavelength is determined from either displacement-distance graph:

• $\lambda =4cm$

To calculate the period, we notice that the difference between the first and second graphs means that the wave has travelled one quarter of a wavelength. This means the period is 4 times the time given between the two displacement-distance graphs.

• T=4(1 s)=4 s

Therefore,

• v=λ/T=4/4=1cm s^−1

Looking at the two graphs, we can also see a crest that is moved 1$cm$ in 1 second, which gives a wave speed of 1 cm s^−1.

Explanation​:

The single slit diffraction effect can not be ignored when slit width is not negligible. Therefore, there will be modulation of the interference pattern. Greater slit widths allow more light reach to screen and makes the fringes brighter. The fringe width is dependent on wavelength, slit separation and distance between screen and slits. None of them have changed. Therefore the fringe width is not affected.

Explanation​:

A greater slit width allows more light to reach the screen. This increases the intensity. So the peaks will be higher. The angular width of maxima is proportional with λ/b. An increase of slit width $b$ results in a decrease in angular width of the maxima. Angular width and width of maxima are proportional. Therefore, the maxima will be narrower.

Explanation​:
Since the nth harmonic has frequency of $n$ times of first harmonic frequency and the ratio of frequencies is 4/2=2. Using $v=f\lambda$ and knowing that the speed of the wave in the string $v$ is unchanged, the ratio of wavelengths is 1/2​.

Explanation​:
C: Correct. In simple harmonic motion, acceleration and displacement always have opposite directions and are proportional

A: Incorrect. Since total energy is the sum of kinetic and potential energies, and the total energy is conserved, so at the time kinetic energy is zero the potential energy must have the maximum value. Since that's not represented in graphs I and II, A is not a correct answer

B: Incorrect. Velocity will be zero at positions of maximum magnitude of acceleration, considering that acceleration and displacement are proportional in magnitude and opposite in direction. So B is not a correct answer.

D: Incorrect The graph of velocity variation with time can be concluded from the slope of displacement variation with time, so the velocity will be zero at positions of maximum magnitude of displacement. Which makes D not a correct answer.

Explanation​:
The speed of a wave depends on the medium, and the frequency of the wave depends on the source.

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.