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Exercises · 14.15

Q.A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency 340 Hz340\ \text{Hz}) when the tube length is 25.5 cm25.5\ \text{cm} or 79.3 cm79.3\ \text{cm}. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.

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The problem uses resonance in a tube closed at one end (piston) and open at the other. The two given lengths correspond to successive resonance modes. The speed of sound is found from the difference in lengths: v=2f(L2−L1)=2×340×(0.793−0.255)≈366 m/sv = 2f(L_2 - L_1) = 2 \times 340 \times (0.793 - 0.255) \approx 366\ \text{m/s}.

A tube with one end closed and the other open supports only odd harmonics of the fundamental. The closed end is a displacement node (pressure antinode), and the open end is a displacement antinode (pressure node). For a given frequency, resonance occurs when the tube length equals an odd multiple of a quarter-wavelength:

L=(2n−1)λ4,n=1,2,3,…L = (2n-1)\frac{\lambda}{4}, \quad n = 1,2,3,\dots

Here the piston acts as the closed end, and the open end is fixed. The tuning fork provides a fixed frequency f=340 Hzf = 340\ \text{Hz}. As the piston is moved, resonance is observed at two different lengths: L1=25.5 cmL_1 = 25.5\ \text{cm} and L2=79.3 cmL_2 = 79.3\ \text{cm}. These must be successive resonance lengths for the same frequency — meaning they correspond to consecutive odd multiples of λ/4\lambda/4.

  1. Identify the mode numbers. Let L1L_1 correspond to n=kn = k and L2L_2 to n=k+1n = k+1 (since they are successive). Then:

L1=(2k−1)λ4,L2=(2(k+1)−1)λ4=(2k+1)λ4L_1 = (2k-1)\frac{\lambda}{4}, \quad L_2 = (2(k+1)-1)\frac{\lambda}{4} = (2k+1)\frac{\lambda}{4}

  1. Subtract to eliminate kk. The difference between the two lengths is:

L2−L1=[(2k+1)−(2k−1)]λ4=2⋅λ4=λ2L_2 - L_1 = \left[(2k+1) - (2k-1)\right]\frac{\lambda}{4} = 2 \cdot \frac{\lambda}{4} = \frac{\lambda}{2}

So the wavelength is twice the difference in lengths:

λ=2(L2−L1)\lambda = 2(L_2 - L_1)

  1. Plug in the numbers. Convert cm to m: L1=0.255 mL_1 = 0.255\ \text{m}, L2=0.793 mL_2 = 0.793\ \text{m}. Then:

λ=2×(0.793−0.255)=2×0.538=1.076 m\lambda = 2 \times (0.793 - 0.255) = 2 \times 0.538 = 1.076\ \text{m}

  1. Use the wave equation. Speed of sound v=fλv = f\lambda: v=340×1.076≈365.84 m/sv = 340 \times 1.076 \approx 365.84\ \text{m/s} …

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