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Q.The fundamental mode of vibration of a stretched string is shown below.

a) Draw the second and third harmonics.
(1)
b) Prove that frequencies produced in the string are in the ratio of 1 : 2 : 3.
(2)
c) Let the fundamental frequency is 45 Hz and the length of the wire is 87.5 cm. If linear density of the wire is 4.0 x 10^-2 kg/m. Find the tension in the string. (2)
Kerala DhseKerala DHSE Plus One Board 2018Subjective· 5mImportance★★★★★
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Figure — Part (a) asks to draw the second and third harmonics of a stretched string fixed at both ends; NCERT Fig 14.13
Figure — Part (a) asks to draw the second and third harmonics of a stretched string fixed at both ends; NCERT Fig 14.13

A string fixed at both ends can only support standing waves with nodes at the two fixed ends, which forces the wavelengths (and hence frequencies) into the ratio 1:2:3:…1:2:3:\dots — this is why plucking a string produces a fundamental plus a full harmonic series. Using T=μ(2Lf1)2T=\mu(2Lf_1)^2 with the given numbers, the tension works out to about 248.1 N.

Setup — the fundamental mode shown in the figure

Figure given: A stretched string fixed at both ends (shown as hatched supports) vibrating in its fundamental mode: a single symmetric lens/eye-shaped loop bulging between the two fixed ends (one antinode at the centre, nodes at the two ends).\text{Figure given: A stretched string fixed at both ends (shown as hatched supports) vibrating in its fundamental mode: a single symmetric lens/eye-shaped loop bulging between the two fixed ends (one antinode at the centre, nodes at the two ends).}

This is the first harmonic (fundamental): one loop, node–antinode–node, wavelength λ1=2L\lambda_1 = 2L.

a) The second and third harmonics

A drawn diagram cannot be rendered in this text answer, but the exact shape each one takes is fully specified below (this is the standard way these are drawn in the NCERT figure, and is sufficient to reproduce them by hand):

  • Second harmonic (n=2n=2): the string vibrates in two equal loops of opposite phase, side by side. There are nodes at both fixed ends and one additional node exactly at the midpoint of the string; antinodes sit at the centre of each loop (at L/4L/4 and 3L/43L/4 from one end). Wavelength λ2=L\lambda_2 = L (one full wavelength fits the string).
  • Third harmonic (n=3n=3): the string vibrates in three equal loops. There are nodes at the two fixed ends plus two additional internal nodes (at L/3L/3 and 2L/32L/3 from one end), with antinodes at the centre of each of the three loops. Wavelength λ3=2L/3\lambda_3 = 2L/3.

In general, the nnth harmonic has nn loops, (n+1)(n+1) nodes total (including the two ends), and nn antinodes.

b) Proof that the frequencies are in the ratio 1 : 2 : 3

A string of length LL fixed rigidly at both ends must have a node at each end — the string cannot move at a rigid support. A standing wave pattern with nn loops fits exactly nn half-wavelengths into the length LL:

L=n⋅λn2⇒λn=2Ln,n=1,2,3,…L = n\cdot\frac{\lambda_n}{2} \quad\Rightarrow\quad \lambda_n = \frac{2L}{n}, \qquad n = 1, 2, 3, \dots

The speed of a transverse wave on the string is set purely by the string's own physical properties — the tension TT and the mass per unit length (linear density) μ\mu — and does not depend on which mode is excited:

v=Tμ(same for every harmonic, since T and μ don’t change)v = \sqrt{\frac{T}{\mu}} \quad \text{(same for every harmonic, since }T\text{ and }\mu\text{ don't change)}

Since v=fnλnv = f_n\lambda_n for every mode, and vv is the same constant for all of them: …

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