A student was studying musical instruments: (a) The student set up a standing wave on a string using the apparatus shown - Edexcel - A-Level Physics - Question 12 - 2023 - Paper 2

The vibrating string produces pressure variations in the air as follows:

1. When the string vibrates, it oscillates back and forth, causing the surrounding air molecules to oscillate as well.
2. This oscillation transfers energy from the vibrating string to the air molecules, generating regions of compression and rarefaction in the air.
3. When the string moves towards a region, it pushes air molecules closer together, creating an area of high pressure (compression). Conversely, when the string moves away, it creates an area of low pressure (rarefaction).
4. These alternating compressions and rarefactions create sound waves that propagate through the air.

To determine if a note of frequency 196 Hz can be played on the string, we first need to find the fundamental frequency based on the tension and mass per unit length. The formula for the fundamental frequency $f$ of a string is given by:

$f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$

where:

• $L$ = length of the string
• $T$ = tension in the string (56 N)
• $\mu$ = mass per unit length of the string ($5.0 \times 10^{-2} kg m^{-1}$)

Considering the maximum vibrating length $L$ = 0.63 m (for 63 cm):

1. Calculate the fundamental frequency:

   • Substituting the values into the formula gives:

   $f = \frac{1}{2 \times 0.63} \sqrt{\frac{56}{5.0 \times 10^{-2}}}$

   • Simplifying this:

   $f = \frac{1}{1.26} \sqrt{1120} \approx \frac{1}{1.26} \times 33.5 \approx 26.5 Hz$

2. Now check with $L$ = 0.21 m (for 21 cm):

   $f = \frac{1}{2 \times 0.21} \sqrt{\frac{56}{5.0 \times 10^{-2}}}$

   • Calculating gives:

   $f = \frac{1}{0.42} \times 33.5 \approx 79.76 Hz$

Thus, the string can play frequencies up to higher harmonics. Since 196 Hz is greater than both calculated frequencies, a note of frequency 196 Hz could be achievable by adjusting the length of the vibrating segment of the string.