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​:
The fringe separation equation is given by s=λ/Dd where $s$ is the fringe separation, λ 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​:
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​:
It is important to locate the mass when t=10s for identifying directions of velocity and acceleration. The time needed for mass to move from one end to the other end is $6s$. The motion of the mass is identical for both sides of the equilibrium position. Therefore, the time needed to cover from one end to equilibrium is $3s$. When mass moves for 10 seconds, it passes the equilibrium position for the second time towards point $X$ and is located between the equilibrium point and point $X$ as shown in the diagram below.

The direction of velocity is towards $X$ (right). The retarding force is always towards equilibrium. Therefore, the acceleration is also towards equilibrium (left).

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.

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​:
The displacements in the diagram shows there are three antinodes and two nodes. This can be represented using a transverse standing wave pattern as shown.

The standing wave is in the second harmonic

For an open pipe, the second harmonic frequency is double that of the first harmonic frequency, therefore statement I is correct.

As the frequency has doubled and the speed of sound has remained the same, through v=fλ, the wavelength will be halved, or λ/2​ which means that statement II in incorrect but statement III is correct.

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.

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.

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​:
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​:

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.

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