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Standing waves on strings IB DP Physics Study Notes

Standing waves on strings IB DP Physics Study Notes - 2025 Syllabus

Standing waves on strings IB DP Physics Study Notes

Standing waves on strings IB DP Physics Study Notes at  IITian Academy  focus on  specific topic and type of questions asked in actual exam. Study Notes focus on IB Physics syllabus with Students should understand

  • nodes and antinodes, relative amplitude and phase difference of points along a standing wave

Standard level and higher level: 4 hours
Additional higher level: There is no additional higher level content .

IB DP Physics 2025 -Study Notes -All Topics

Boundary conditions – strings

  • You may be wondering how a situation could ever develop in which two identical waves come from opposite directions. Well, wonder no more.
  • When you pluck a stringed instrument, waves travel to the ends of the string and reflect at each end, and return to interfere under precisely the conditions needed for a standing wave.
  • Note that there are two nodes and one antinode.
  • Why must there be a node at each end of the string?

Sketching and interpreting standing wave patterns

        

  • Observe that precisely half a wavelength fits along the length of the string or $\frac{1}{2}\lambda = L$.
  • Thus we see that $\lambda = 2L$.
  • Since $v = f\lambda$ we see that $f = \frac{v}{2L}$ for a string.
  • This is the lowest frequency you can possibly get from this string configuration, so we call it the fundamental frequency $f_1$.
  • The fundamental frequency of any system is also called the first harmonic.

  •  The next higher frequency has another node and another antinode.
  • We now see that λ = L.
  • Since v = λf we see that f = v/L.
  • This is the second lowest frequency you can possibly get and since we called the fundamental frequency f₁, we’ll name this one f₂.
  • This frequency is also called the second harmonic.

Example:

A guitar string of length \( L = 0.65 \, \text{m} \) is fixed at both ends. The wave speed on the string is \( v = 520 \, \text{m/s} \).

(a) Calculate the frequency of the fundamental mode.

(b) Determine the wavelength and frequency of the second harmonic.

▶️ Answer/Explanation

(a) Fundamental Frequency:

For the first harmonic (fundamental), \( \lambda_1 = 2L = 2 \times 0.65 = 1.30 \, \text{m} \)
Using \( f = \frac{v}{\lambda} \Rightarrow f_1 = \frac{520}{1.30} = \boxed{400 \, \text{Hz}} \)

(b) Second Harmonic:

For the second harmonic, \( \lambda_2 = L = 0.65 \, \text{m} \)
\( f_2 = \frac{v}{\lambda_2} = \frac{520}{0.65} = \boxed{800 \, \text{Hz}} \)

IB Physics Standing waves on strings Exam Style Worked Out Questions

Question

A standing wave is formed on a string. P and Q are adjacent antinodes on the wave. Three   statements are made by a student:
                        I. The distance between P and Q is half a wavelength.
                        II. P and Q have a phase difference of π rad.
                        III. Energy is transferred between P and Q.
     Which statements are correct?
     A. I and II only
     B. I and III only
     C. II and III only
     D. I, II and III

Answer/Explanation

Ans: A

There is no transport of energy in standing wave because the two waves that make them up carry equal energy in opposite directions.

Question

A string stretched between two fixed points sounds its second harmonic at frequency f.

Which expression, where n is an integer, gives the frequencies of harmonics that have a node at the centre of the string?

A. \(\frac{{n + 1}}{2}f\)

B. nf

C. 2nf

D. (2n + 1)f

Answer/Explanation

Markscheme

B

\(n\frac{\lambda}{2}=L\)
\(\therefore f_n =\frac{v}{\lambda} =\frac{nv}{2L}\) —(1)
now for first Harmonic \(n=1\)
or
\(\lambda = 2L\)
\(\therefore f=\frac{v}{\lambda}=\frac{v}{2L}\)
putting this value in equation (1)
\(f_n =\frac{nv}{2L} =nf\)

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