(a) (i) state two characteristics of resonance.
(ii) mention the effects of temperature and pressure on the speed of sound.
(iii) list the types of resonance as applied to wave. 7 marks
(b) a referee leaning on a wall blows his whistle towards another wall 50 m away. he hears the second echo 2 seconds later. calculate the velocity of the sound of the whistle in air. speed of sound in air = 330 ms⁻¹ 3 marks
(c) (i) draw a diagram illustrating convergent and divergent beams of light.
(ii) the critical angle for glass in air is 39°. calculate the refractive index of the glass.

Question

Accepted Answer

### Part (a)
#### (i) Characteristics of Resonance
- **Explanation**: Resonance occurs when a system is driven at its natural frequency. Two key characteristics are:
    - 1. A large amplitude of vibration is produced when the driving frequency matches the natural frequency of the system.
    - 2. The phase difference between the driving force and the displacement of the oscillating system is $90^{\circ}$ (or $\frac{\pi}{2}$ radians) at resonance.
#### (ii) Effects of Temperature and Pressure on Speed of Sound
- **Temperature**: The speed of sound in a gas (like air) is directly proportional to the square root of the absolute temperature. Mathematically, $v \propto \sqrt{T}$, where $v$ is the speed of sound and $T$ is the absolute temperature. As temperature increases, the speed of sound increases because the molecules of the medium vibrate more vigorously, leading to faster transfer of sound energy.
- **Pressure**: For an ideal gas, the speed of sound is independent of pressure when the temperature is constant. This is because, according to the ideal gas law, if pressure increases while temperature remains constant, the density of the gas also increases proportionally, and the effect of pressure increase on speed (which is related to $\sqrt{\frac{P}{
ho}}$) cancels out since $P$ and $
ho$ change in the same ratio.
#### (iii) Types of Resonance (Wave - related)
- 1. **Mechanical Resonance**: Occurs in mechanical systems like a vibrating string, a pendulum, or a tuning fork. For example, a swing (pendulum - like system) resonates when pushed at its natural frequency.
- 2. **Acoustic Resonance**: Related to sound waves. For example, resonance in a pipe (like organ pipes) where standing sound waves are formed when the frequency of the sound source matches the natural frequency of the pipe.
- 3. **Electrical Resonance**: Occurs in electrical circuits (like LC circuits) where the inductive reactance equals the capacitive reactance, leading to a maximum current (in series resonance) or minimum impedance (in parallel resonance). But for wave - related resonance, mechanical and acoustic are more relevant. So, two types can be mechanical resonance (in mechanical wave systems) and acoustic resonance (in sound wave systems).

### Part (b)
# Explanation:
## Step 1: Determine the total distance traveled by sound
The referee blows the whistle towards a wall 50 m away. For the second echo, the sound has to travel to the wall and back to the referee, and then again to the wall and back? Wait, no. Wait, when the referee blows the whistle, the first echo would be from the near wall? But the problem says "towards another wall 50 m away" and "hears the second echo 2 seconds later". Wait, let's think again. The sound travels to the far wall (50 m) and back to the referee, that's a distance of $d_1=50\times2 = 100$ m for the first echo? No, maybe the second echo: the sound goes to the far wall (50 m), comes back to the referee (50 m), then goes to the near wall (let's assume the referee is between two walls? Wait, the problem says "a referee leaning on a wall blows his whistle towards another wall 50 m away". So there are two walls: the one he is leaning on (wall A) and another wall (wall B) 50 m away. When he blows the whistle, the sound travels to wall B (50 m), reflects back to wall A (50 m), then reflects back to wall B? No, the second echo: the time for the second echo is the time for the sound to go to wall B, come back to wall A, then go to wall B again? Wait, no. Let's re - read: "He hears the second echo 2 seconds later". The first echo would be when the sound goes to wall B and comes back to him (distance $2\times50 = 100$ m). The second echo would be when the sound goes to wall B, comes back to him, then goes to wall A (the wall he is leaning on), reflects, and comes back to him? Wait, no, maybe the problem is simpler. Wait, the referee is leaning on a wall (so his distance from that wall is 0). He blows the whistle towards another wall 50 m away. The sound travels to the 50 - m wall (distance 50 m), reflects, and comes back to him (distance 50 m) for the first echo. For the second echo, the sound would go from him to the 50 - m wall (50 m), reflect to his wall (0 - distance wall, 50 m), reflect again to the 50 - m wall (50 m), and then reflect back to him (50 m)? No, that seems complicated. Wait, maybe the total distance for the second echo is $2\times50\times2=200$ m? Wait, the time given is 2 seconds. Let's use the formula $v=\frac{d}{t}$.

Wait, when the referee blows the whistle, the sound travels to the wall 50 m away (distance $d_1 = 50$ m) and back to him (distance $d_1=50$ m) for the first echo. For the second echo, the sound has to travel to the wall, back to him, then to the wall again, and back to him? No, maybe the problem is that the second echo is the echo that has traveled to the wall and back twice? Wait, no. Let's think of the path: the referee is at position O (leaning on wall 1), wall 2 is 50 m from O. The sound from O to wall 2 (50 m), wall 2 to O (50 m) - first echo. Then, the sound from O to wall 2 (50 m), wall 2 to wall 1 (50 m), wall 1 to wall 2 (50 m), wall 2 to O (50 m) - second echo. So the total distance for the second echo is $50 + 50+50 + 50=200$ m? And the time taken is 2 seconds.

## Step 2: Calculate the velocity
We know that speed $v=\frac{\text{distance}(d)}{\text{time}(t)}$. The total distance traveled by sound for the second echo is $d = 2\times50\times2=200$ m (because it goes to the wall, comes back, goes to the wall again, comes back? Wait, no, maybe the first echo is at time $t_1=\frac{2\times50}{v}$, and the second echo is at time $t_2=\frac{4\times50}{v}$. The difference between $t_2$ and $t_1$ is 2 seconds. So $\frac{4\times50}{v}-\frac{2\times50}{v}=2$. Simplify: $\frac{200 - 100}{v}=2$, $\frac{100}{v}=2$, so $v = 50$ m/s? Wait, that can't be right, because the speed of sound in air is given as 330 m/s. Wait, maybe I misinterpret the problem.

Wait, the problem says "a referee leaning on a wall blows his whistle towards another wall 50 m away. He hears the second echo 2 seconds later." Let's consider that the first echo is when the sound travels to the 50 - m wall and back (distance $2\times50 = 100$ m). The second echo is when the sound travels to the 50 - m wall, back to the referee, then to the wall he is leaning on (distance 0, so no, that wall is behind him), wait, maybe the referee is between two walls: wall A (he is leaning on it) and wall B (50 m from wall A, so 50 m from him). So when he blows the whistle, the sound travels to wall B (50 m), reflects to wall A (50 m), reflects to wall B (50 m), and then reflects to him (50 m). So the total distance for the second echo is $50+50 + 50+50 = 200$ m, and the time taken is 2 seconds. Then $v=\frac{200}{2}=100$ m/s? No, that's still not 330. Wait, maybe the problem is that the "second echo" is the echo that has traveled to the wall and back once, but the first echo is from the near wall? Wait, the referee is leaning on a wall, so the near wall is at 0 distance. So when he blows the whistle, the sound travels to the far wall (50 m) and back (50 m) - that's the first echo (distance 100 m). Then, the sound travels to the near wall (0 m, so no distance), reflects, and travels to the far wall (50 m) and back (50 m) - that's the second echo. Wait, no, the near wall is where he is leaning, so the sound can't travel to the near wall and back because he is on it. I think there is a misinterpretation. Let's re - read the problem: "A referee leaning on a wall blows his whistle towards another wall 50 m away. He hears the second echo 2 seconds later."

Alternative approach: The sound has to travel to the wall (50 m) and back to the referee, and then again to the wall and back? No, the time between the first and second echo is 2 seconds. Wait, maybe the first echo is when the sound goes to the wall and back (distance $2\times50$ m), and the second echo is when the sound goes to the wall, back to the referee, then to the wall again, and back to the referee. So the distance for the second echo (from the time of blowing) is $4\times50$ m, and the time for the second echo is $t=\frac{4\times50}{v}$, and the time for the first echo is $t_1=\frac{2\times50}{v}$. The difference $t - t_1=2$ seconds. So $\frac{200}{v}-\frac{100}{v}=2$, $\frac{100}{v}=2$, $v = 50$ m/s. But this contradicts the given speed of sound (330 m/s). Maybe the problem has a typo, or my interpretation is wrong. Wait, maybe the "second echo" is the echo that travels to the wall and back once, but the first echo is from a different source? No, the problem says "the second echo", so the first echo is when the sound goes to the wall and back (distance 100 m), and the second echo is when the sound goes to the wall, back to the referee, then to the wall again, and back to the referee (distance 200 m), and the time between the first and second echo is 2 seconds. So the time for the second echo (from blowing) is $t_2=\frac{200}{v}$, and the time for the first echo is $t_1=\frac{100}{v}$, so $t_2 - t_1=2$, $\frac{200 - 100}{v}=2$, $v = 50$ m/s. But this is inconsistent with the given 330 m/s. Maybe the problem means that the time taken for the second echo (from blowing) is 2 seconds. So the sound travels to the wall (50 m) and back to the referee (50 m), and then to the wall again (50 m) and back to the referee (50 m), total distance $4\times50 = 200$ m, time $t = 2$ s, so $v=\frac{200}{2}=100$ m/s. Still not 330. Maybe the problem is that the "second echo" is the echo that travels to the wall and back once, and the time is 2 seconds. So distance $d = 2\times50=100$ m, time $t = 2$ s, $v=\frac{100}{2}=50$ m/s. But this is wrong. Wait, maybe the referee is 50 m from the wall, and the second echo is the echo that has bounced off the wall twice. So the sound travels 50 m to the wall, 50 m back to the referee, 50 m to the wall, 50 m back to the referee: total distance $4\times50 = 200$ m, time 2 s, so $v = 100$ m/s. I think there is a mistake in the problem, but following the calculation:

## Step 1: Calculate total distance
The sound travels to the wall (50 m) and back to the referee (50 m) for the first echo. For the second echo, it travels to the wall, back to the referee, to the wall, and back to the referee. So total distance $d=4\times50 = 200$ m.

## Step 2: Calculate velocity
Using $v=\frac{d}{t}$, where $d = 200$ m and $t = 2$ s. So $v=\frac{200}{2}=100$ m/s. But the given speed of sound is 330 m/s, so maybe the problem's "second echo" is the first echo, and there is a misstatement. Alternatively, if we consider that the sound travels to the wall and back once (distance 100 m) in 2 seconds, then $v=\frac{100}{2}=50$ m/s. But this is not correct. Maybe the problem is that the referee is 50 m from the wall, and the echo is heard 2 seconds later, so the sound travels 50 m to the wall and 50 m back, total 100 m, time 2 s, so $v=\frac{100}{2}=50$ m/s. But this is inconsistent with the given 330 m/s. Maybe the problem has a typo, and the time is 0.3 s (approx) for 330 m/s. Anyway, following the problem's numbers:

# Answer (Part b):
The velocity of sound is $\boldsymbol{100\space m/s}$ (or 50 m/s depending on interpretation, but based on the second echo traveling 200 m in 2 s, it's 100 m/s).

### Part (c)
#### (i) Diagram of Convergent and Divergent Beams of Light
- **Convergent Beam**: Rays of light that are coming from different directions and are moving towards a common point (the focus). For example, rays from a distant object converging towards the focal point of a convex lens.
- **Divergent Beam**: Rays of light that are coming from a common point and are spreading out. For example, rays coming out of a point source (like a bulb) or rays coming out of a concave lens.

A simple diagram:

- For convergent beam: Draw several lines (rays) that start from different points on the left and meet at a point (focus) on the right.
- For divergent beam: Draw several lines (rays) that start from a common point (source) on the left and spread out as they move to the right.

#### (ii) Calculation of Refractive Index
# Explanation:
## Step 1: Recall the formula for critical angle
The critical angle $C$ is related to the refractive index $n$ of the denser medium (glass) with respect to the rarer medium (air) by the formula $\sin C=\frac{1}{n}$ (when the rarer medium is air, $n_{air}\approx1$).

## Step 2: Solve for refractive index
We are given $C = 39^{\circ}$. From $\sin C=\frac{1}{n}$, we can re - arrange to get $n=\frac{1}{\sin C}$.

Calculate $\sin(39^{\circ})\approx0.6293$.

Then $n=\frac{1}{0.6293}\approx1.59$.

# Answer (Part c ii):
The refractive index of the glass is approximately $\boldsymbol{1.59}$ (or more accurately, using a calculator, $\sin(39^{\circ})\approx0.6293$, so $n=\frac{1}{0.6293}\approx1.59$).

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QUESTION IMAGE

Question

Explanation:

Part (a)

(i) Characteristics of Resonance

(ii) Effects of Temperature and Pressure on Speed of Sound

(iii) Types of Resonance (Wave - related)

Part (b)

Step 1: Determine the total distance traveled by sound

Step 2: Calculate the velocity

Step 1: Recall the formula for critical angle

Step 2: Solve for refractive index

Answer:

Part (c)

(i) Diagram of Convergent and Divergent Beams of Light

(ii) Calculation of Refractive Index