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PHYSICS, M3 2023 VCE 11

A guitar string of length 0.75 m and fixed at both ends is plucked and a standing wave is produced. The envelope of the standing wave is shown in Figure 13.
 

The speed of the wave along the string is 393 m s\( ^{-1}\).

  1. What is the frequency of the wave?  (1 mark)

  2. Describe how the standing wave is produced on the string fixed at both ends.  (2 marks)

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a.    262 Hz

b.   Standing wave:

→ When waves encounter fixed ends, they reflect.

→ If the string’s length is a multiple of half the wavelength, the reflected wave combines with the original wave, resulting in interference that forms a standing wave pattern

Show Worked Solution

a.   \(f=\dfrac{v}{\lambda}=\dfrac{393}{1.5}=262\ \text{Hz}\)
 

b.   Standing wave:

→ When waves encounter fixed ends, they reflect.

→ If the string’s length is a multiple of half the wavelength, the reflected wave combines with the original wave, resulting in interference that forms a standing wave pattern

♦ Mean mark (b) 47%.

PHYSICS, M3 2017 VCE 16

Standing waves are formed on a string of length 4.0 m that is fixed at both ends. The speed of the waves is 240 m s\(^{-1}\).

  1. Calculate the wavelength of the lowest frequency resonance.  (2 marks)

  2. Calculate the frequency of the second-lowest frequency resonance.  (2 marks)

  3. Explain the physics of how standing waves are formed on the string. Include a diagram in your response.  (3 marks)

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a.    30 Hz

b.    60 Hz

c.    Standing waves:

→ Produced when a wave travels down a string and is reflected back on itself such that the superposition of the two waves produce an interference pattern to form a standing wave.

→ The two waves must be travelling in opposite directions with the same frequency, wavelength and amplitude.


       

Show Worked Solution

a.     Lowest frequency resonance:

→ Occurs at the maximum wavelength. The maximum wavelength is 8 metres since the half wavelength is the length of the string (4m).

\(f=\dfrac{v}{\lambda}=\dfrac{240}{8}=30\ \text{Hz}\)
 

b.     When \(\lambda = 4: \)

\(f=\dfrac{v}{\lambda}=\dfrac{240}{4}=60\ \text{Hz}\)
 

c.    Standing waves:

→ Produced when a wave travels down a string and is reflected back on itself such that the superposition of the two waves produce an interference pattern to form a standing wave.

→ The two waves must be travelling in opposite directions with the same frequency, wavelength and amplitude.


       

♦ Mean mark (c) 43%.

PHYSICS, M3 2018 VCE 11

Figure 13 shows two speakers, \(\text{A}\) and \(\text{ B}\), facing each other. The speakers are connected to the same signal generator/amplifier and the speakers are simultaneously producing the same 340 Hz sound.
 

Take the speed of sound to be 340 ms\( ^{-1}\).

  1. Calculate the wavelength of the sound.  (1 mark)

  2. A student stands in the centre, equidistant from speakers \(\text{A}\) and \(\text{B}\). He then moves towards speaker \(\text{B}\) and experiences a sequence of loud and quiet regions. He stops at the second region of quietness.
  3. How far has the student moved from the centre? Explain your reasoning.  (3 marks)

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a.     \(\lambda=1\ \text{m}\)

b.    → At the centre of the speakers, there is an area of relative loudness

→ This is due to a path difference of 0 metres between the wavelengths of sounds coming from both speaker \(\text{A}\) and speaker \(\text{B}\). 

→ As the student moves towards speaker \(\text{B}\), the two sound waves will experience constructive (loud) and destructive interference (quiet).

→ The peak of speaker \(\text{A}\) will move into the trough of speaker \(\text{B}\) where he experiences the first region of quietness (node) which is 1 quarter wavelength (0.25m) from the centre.

→ The second region of quietness (node) will be half of a wavelength (0.5m) from the first region of quietness. Thus, the total distance from the centre to the second region of quietness is 0.75m.

Show Worked Solution

a.    \(\lambda=\dfrac{v}{f}=\dfrac{340}{340}=1\ \text{m}\)

 

b.    → At the centre of the speakers, there is an area of relative loudness

→ This is due to a path difference of 0 metres between the wavelengths of sounds coming from both speaker \(\text{A}\) and speaker \(\text{B}\). 

→ As the student moves towards speaker \(\text{B}\), the two sound waves will experience constructive and destructive interference.

→ The peak of speaker \(\text{A}\) will move into the trough of speaker \(\text{B}\) where he experiences the first region of quietness (node) which is 1 quarter wavelength (0.25m) from the centre.

→ The second region of quietness (node) will be half of a wavelength (0.5m) from the first region of quietness. Thus, the total distance from the centre to the second region of quietness is 0.75m.

♦♦♦ Mean mark (b) 20%.

PHYSICS, M3 2019 VCE 12

A sinusoidal wave of wavelength 1.40 m is travelling along a stretched string with constant speed \(v\), as shown in the figure below. The time taken for point \(\text{P}\) on the string to move from maximum displacement to zero is 0.120 s.
 

Calculate the speed of the wave, \(v\). Give your answer correct to three significant figures. Show your working.  (3 marks)

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\(v=2.92\ \text{ms}^{-1}\)

Show Worked Solution

→ Using  \(v=f\lambda\) and  \(f=\dfrac{1}{T}\):

\(v\) \(=\dfrac{\lambda}{T}\)  
  \(=\dfrac{1.4}{4 \times 0.120}\)  
  \(=2.92\ \text{ms}^{-1}\)  
♦ Mean mark 50%.

PHYSICS, M3 2021 VCE 13

The diagram below shows part of a travelling wave.
 

The wave propagates with a speed of 18 ms\(^{-1}\).

What is the amplitude and frequency of the wave?   (2 marks)

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Amplitude = 8 cm

Frequency = 300 Hz

Show Worked Solution

→ Amplitude: the distance from the centre line to maximum displacement.

→ Amplitude = 8 cm.

→ Frequency: can be determined using  \(f=\dfrac{v}{\lambda}\), (wavelength must be in metres)

\(f=\dfrac{v}{\lambda}=\dfrac{18}{0.06}=300\ \text{Hz} \)

PHYSICS, M3 2022 VCE 13*

A travelling wave produced at point \(\text{A}\) is reflected at point \(\text{B}\) to produce a standing wave on a rope, as represented in the diagram below.
 

 

The distance between points \(\text{A}\) and \(\text{B}\) is 2.4 m. The period of vibration of the standing wave is 1.6 s.

Find the speed of the travelling wave along the rope in metres per second.   (2 marks)

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\(A\)

Show Worked Solution

→ \( \lambda = 1.2\) m

→ \(f= \dfrac{1}{T} = \dfrac{1}{1.6} = 0.625\)

\(v\) \(=f \lambda\)  
  \(=0.625 \times 1.2\)  
  \(=0.75\ \text{ms}^{-1}\)  

PHYSICS, M3 2017 HSC 21a

A laser emits light of wavelength 550 nm. (\(v=3 \times 10^8\) ms\(^{-1}\))

Calculate the frequency of this light.   (2 marks)

 

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\(f= 5.45 \times 10^{14}\) Hz

Show Worked Solution
\(v\) \(=f \lambda\)  
\(f\) \(=\dfrac{v}{\lambda}\)  
\(f\) \(=\dfrac{3 \times 10^8}{550 \times 10^{-9}}\)  
\(f\) \(=5.45 \times 10^{14}\) Hz