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​:

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​:
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​:
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 x^2 in the expression. These two conditions are satisfied in B.

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​:
Intensity falls off significantly from the principal maximum to secondary maxima. It decreases to approximately 5 % of the initial value. The exact location of $X$ is not known. The maximum intensity that $X$ can have is 5 % of the initial value. Therefore, the ratio should be equal to or less than 0.05.

Explanation​:
The fringe separation equation is given by s=λD/d​ where $s$ is the fringe separation, $\lambda$ is the wavelength, $D$ is the distance between the double slit barrier and the screen and s is the distance between the two slits on the double slit barrier. The question is asking which change will increase the value of s.

• A. Moving the screen closer to the double slit barrier will decrease D. Since $D$ is directly proportional to $s$, $s$ will also decrease.
• B. Violet light has a higher frequency / shorter wavelength than blue light. Using violet light instead of blue light will decrease the value of $\lambda$, and since $\lambda$ is also directly proportional to $s$, $s$ will subsequently decrease.
• C. Since $d$, the space between the two slits on the double slit barrier, is inversely proportional to $s$, as $d$ decreases, $s$ will increase.
• D. As long as the aperture is not completely closed, it will have no bearing on the spacing of the double slit interference pattern and will only reduce its intensity. The purpose of the single slit is to produce coherent light through diffraction, which then falls on the double slits.

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.

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​:

The angular width of the first maximum;

• θ=λ/b​

The central maximum has the twice angular width of the first maximum;

• θc=2λ/b​

Since wavelength λ and slit width b are not changed, the angular width must be constant. Therefore, the answer is C.

Given that the angular width is constant, increasing the distance D will cause the size of the central maximum on the screen to double, which indicates that option A is incorrect.

Explanation​:
Since frequency is 0.05 Hz, T=1/f=1/0.05=20 s

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​:
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.

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

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.

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.