Stationary wavesEdexcel A-Level Physics: Subtopic test

1. 1
   A taut wire of length 0.60 m is fixed at both ends and plucked so that it vibrates with a single loop. The wire has a mass per unit length of 4.0 × 10⁻⁴ kg m⁻¹ and is under a tension of 36 N. Treat both fixed ends as nodes.

   (a)

   What is the speed of transverse waves on the wire?

   [1 mark]

   • A9.0 × 10⁴ m s⁻¹
   • B300 m s⁻¹
   • C0.12 m s⁻¹
   • D1.5 × 10⁴ m s⁻¹

   (b)

   What is the frequency of the fundamental vibration of the wire?

   [1 mark]

   • A500 Hz
   • B125 Hz
   • C300 Hz
   • D250 Hz

   (c)

   The tension is increased to 144 N with the length unchanged. Calculate the new fundamental frequency.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   A string is stretched between two fixed points 1.50 m apart. A vibration generator near one end drives the string at 120 Hz, and a stationary wave with exactly three loops is seen between the fixed ends. Treat both ends as nodes.

   (a)

   Which statement describes a node of the stationary wave?

   [1 mark]

   • AA point of zero amplitude where the two waves are always in antiphase
   • BA point of maximum amplitude where the waves are in phase
   • CA point where the string moves with the wave speed
   • DA point where the string has maximum amplitude only at certain times

   (b)

   What is the distance between two adjacent nodes?

   [1 mark]

   • A1.00 m
   • B0.75 m
   • C0.50 m
   • D0.25 m

   (c)

   Explain how the stationary wave is formed on the string.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   A student investigates how the fundamental frequency f of a vibrating string depends on its tension T. A string of vibrating length 0.800 m passes over a pulley and carries hanging masses that provide the tension. For each tension she adjusts the frequency of a vibration generator until the string shows a single loop of maximum amplitude. She plots f² against T and obtains a straight line through the origin with gradient 391 Hz² N⁻¹.

   (a)

   Use the gradient to determine the mass per unit length of the string.

   [3 marks]

   (b)

   Explain why the student plots f² against T rather than f against T, and suggest two ways of improving the accuracy of her measurements of the fundamental frequency.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   A guitar string of length 0.50 m is fixed at both ends. It has a mass per unit length of 5.0 × 10⁻³ kg m⁻¹ and is under a tension of 50 N.

   (a)

   Explain how a stationary wave is formed when the string is plucked, why the string can vibrate only at certain frequencies, and calculate the fundamental frequency.

   [6 marks]

   (b)

   The guitarist wants to raise the fundamental frequency of this string from 100 Hz to 125 Hz. She can either press the string against a fret to shorten its vibrating length, or tighten the string. Evaluate these two methods, using calculations.

   [6 marks]

   Total for question 4: 12 marks

End of questions