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CalibratePhysics

6.8

Sonometer

Superposition of Waves · Maharashtra HSC

Theory

Construction

A sonometer consists of a hollow rectangular wooden box called the sound box. The sound box makes a larger mass of air vibrate, so the sound produced by the vibrating metal wire is amplified; the same principle is used in stringed instruments such as the violin, the guitar and the tanpura.

Two bridges P and Q run across the width of the box and can be moved parallel to its length. A metal wire of uniform cross-section runs along the length of the box over these bridges. It is fixed at one end, and its other end passes over a pulley and carries a hanger to which slotted weights can be added. Changing the weights changes the tension in the wire; moving the bridges changes the vibrating length AB.

Working

If the wire is plucked at a point midway between the bridges, transverse waves are produced. They are reflected at the bridges and superpose, so stationary waves are set up between the two bridges; the portion AB of the wire between P and Q is the vibrating length.

The wire can also be made to vibrate by holding a vibrating tuning fork near it, or by holding the stem of the fork against the sonometer box. The frequency of vibration is then the same as that of the fork. If this frequency happens to be one of the natural frequencies of the wire, standing waves of large amplitude are set up because the two vibrate in resonance.

To identify resonance, a small piece of paper called the rider R is placed over the wire midway between the bridges. When the frequency of the tuning fork matches the frequency of the fundamental mode of the wire — achieved by adjusting the length AB with the bridges P and Q — the rider sits at an antinode and is thrown off the wire.

Diagram in Words

Draw a long hollow rectangular box seen from the side. A wire runs along the top, fixed to a peg at the left end, passing over two inverted-V bridges P and Q, then over a pulley clamped at the right edge of the table, carrying a hanger with slotted weights. Mark the vibrating length AB between the bridges and place a small paper rider R at the midpoint of AB. Label the box as the sound box.

Verification of the Laws of a Vibrating String

First law — the law of length

By measuring the length of the wire and its mass, the mass per unit length mm is determined. The wire is then stretched on the sonometer and the hanger is suspended from its free end. A suitable tension TT is applied by placing slotted weights on the hanger.

A light paper rider is placed on the wire midway between the bridges. The tuning fork is set into vibration by striking it on a rubber pad, and its stem is held in contact with the sonometer box. By changing the distance between the bridges without disturbing the rider, the frequency of vibration of the wire is changed. When the frequency of the wire becomes exactly equal to that of the fork, the wire vibrates with maximum amplitude and the paper rider is thrown off; that length is noted.

A set of tuning forks of different frequencies n1,n2,n3,…n_1, n_2, n_3,\ldots is used, and the corresponding vibrating lengths l1,l2,l3,…l_1, l_2, l_3,\ldots are noted, keeping the tension TT constant. It is observed that

n1l1=n2l2=n3l3=…=constantn_1l_1 = n_2l_2 = n_3l_3 = \ldots = \text{constant}

∴ nl=constanti.e.n∝1lif T and m are constant\therefore\ nl = \text{constant} \qquad \text{i.e.}\quad n \propto \dfrac{1}{l} \quad \text{if } T \text{ and } m \text{ are constant}

Thus the first law of a vibrating string is verified.

Second law — the law of tension

The vibrating length ll of the given wire of mass per unit length mm is kept constant. By changing the tension, the same length is made to vibrate in unison with different tuning forks of various frequencies. If tensions T1,T2,T3,…T_1, T_2, T_3,\ldots correspond to frequencies n1,n2,n3,…n_1, n_2, n_3,\ldots it is observed that

n1T1=n2T2=n3T3=…=constant\dfrac{n_1}{\sqrt{T_1}} = \dfrac{n_2}{\sqrt{T_2}} = \dfrac{n_3}{\sqrt{T_3}} = \ldots = \text{constant}

∴ nT=constanti.e.n∝Tif l and m are constant\therefore\ \dfrac{n}{\sqrt{T}} = \text{constant} \qquad \text{i.e.}\quad n \propto \sqrt{T} \quad \text{if } l \text{ and } m \text{ are constant}

Third law — the law of linear density

Two wires of different mass per unit length m1m_1 and m2m_2 are used. The first wire is given a suitable tension and made to vibrate in unison with a given tuning fork; its vibrating length l1l_1 is noted. Using the same fork, the second wire is made to vibrate under the same tension and its vibrating length l2l_2 is determined. The frequency of vibration of the two wires is thus the same, under the same applied tension TT. It is found that

l1m1=l2m2∴ lm=constantl_1\sqrt{m_1} = l_2\sqrt{m_2} \qquad \therefore\ l\sqrt{m} = \text{constant}

But by the first law n∝1ln \propto \dfrac{1}{l}, therefore

n∝1mif T and l are constantn \propto \dfrac{1}{\sqrt{m}} \quad \text{if } T \text{ and } l \text{ are constant}

In this way all three laws of a vibrating string are verified using a sonometer.

Load Immersed in Water

If the load stretching the wire has specific gravity ρ\rho and is completely immersed in water, the tension falls from w1w_1 to the apparent weight w2w_2, with w1w1−w2=ρ\dfrac{w_1}{w_1 - w_2} = \rho, so w1w2=ρρ−1\dfrac{w_1}{w_2} = \dfrac{\rho}{\rho - 1}. Since n∝Tn \propto \sqrt{T},

n1n2=w1w2=ρρ−1\dfrac{n_1}{n_2} = \sqrt{\dfrac{w_1}{w_2}} = \sqrt{\dfrac{\rho}{\rho - 1}}

The frequency falls, so the beats heard with the fork give n1−n2n_1 - n_2.

Standard Results

  • Resonance is detected by the paper rider flying off the wire.
  • Law of length: nl=nl = constant. Law of tension: nT=\dfrac{n}{\sqrt{T}} = constant. Law of linear density: lm=l\sqrt{m} = constant.
  • n=12lTmn = \dfrac{1}{2l}\sqrt{\dfrac{T}{m}}, with T=MgT = Mg for a hanging mass MM and g=9.8g = 9.8 m s−2^{-2}.
  • Load of specific gravity ρ\rho immersed in water: n1n2=ρρ−1\dfrac{n_1}{n_2} = \sqrt{\dfrac{\rho}{\rho - 1}}.

Theory Questions Asked on This Subtopic

Real board questions. Answers are at the end of the page.

  1. 1

    Explain how law of length can be verified by using a sonometer.

    Maharashtra HSC board question — asked in October 2009Answer 1 ↓

Numericals

Every numerical set on this subtopic. Full working is at the end of the page.

  1. 2

    A sonometer wire is in unison with a tuning fork when stretched by a weight of specific gravity 'nine'. On completely increasing the weight in water, wire produces 4 beats per second with the fork. Calculate the frequency of the fork.

    Maharashtra HSC board question — asked in March 2009Answer 2 ↓
  2. 3

    A 36 cm long sonometer wire vibrates with frequency of 280 Hz in fundamental mode, when it is under tension of 24.5 N. Calculate linear density of the material of wire.

    Maharashtra HSC board question — asked in October 2014Answer 3 ↓
  3. 4

    A sonometer wire is in unison with a tuning fork of frequency 125 Hz when it is stretched by a weight. When the weight is completely immersed in water, 8 beats are heard per second. Find the specific gravity of the material of the weight.

    Maharashtra HSC board question — asked in October 2015Answer 4 ↓
  4. 5

    A stretched sonometer wire is in unison with a tuning fork. When the length of the wire is increased by 5%, the number of beats heard per second is 10. Find the frequency of the tuning fork.

    Maharashtra HSC board question — asked in March 2019Answer 5 ↓
  5. 6

    A sonometer wire 1 metre long weighing 2 g is in resonance with a tuning fork of frequency 300 Hz. Find tension in the sonometer wire.

    Maharashtra HSC board question — asked in March 2019Answer 6 ↓

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

  1. Verification of first law: i. By measuring length of wire and its mass, the mass per unit length (m) of wire is determined. Then the wire is stretched on the sonometer and the hanger is suspended from its free end. ii. A suitable tension (T) is applied to the wire by placing slotted weights on the hanger. iii. The length of wire (l1)(l_1) vibrating with the same frequency (n1)(n_1) as that of the tuning fork is determined as follows. iv. A light paper rider is placed on the wire midway between the bridges. The tuning fork is set into vibrations by striking on a rubber pad. v. The stem of tuning fork is held in contact with the sonometer box. By changing distance between the bridges without disturbing paper rider, frequency of vibrations of wire is changed. vi. When the frequency of vibrations of wire becomes exactly equal to the frequency of tuning fork, the wire vibrates with maximum amplitude and the paper rider is thrown off. vii. In this way a set of tuning forks having different frequencies n1,n2,n3,…n_1,n_2,n_3,…are used and corresponding vibrating lengths of wire are noted as l1,l2,l3……..l_1,l_2,l_3……..by keeping the tension constant (T). viii. It is observed that n1l1=n2l2=……=n_1l_1=n_2l_2=……= constant, for constant value of tension (T) and mass per unit length(m). ∴nl=\therefore nl= constant i.e., n∝1ln\propto\dfrac{1}{l}, if T and m are constant. Thus, the first law of a vibrating string is verified by using a sonometer.

  2. Given: Specific gravity = 9, Beat frequency = 4 Hz To find: Frequency (n1)(n_1) Formula: n=12lTmn=\dfrac{1}{2l}\sqrt{\dfrac{T}{m}} Calculation: Let, w1w_1 = weight of body in air, w2w_2 = apparent weight of body in water, n1n_1 = frequency of fork Specific gravity, ρ=w1w1−w2=9\rho=\dfrac{w_1}{w_1-w_2}=9 ∴w1w2=98\therefore \dfrac{w_1}{w_2}=\dfrac{9}{8} From formula, Fundamental frequency of vibration of wire, n=12lwmn=\dfrac{1}{2l}\sqrt{\dfrac{w}{m}} ∴n1=12lw1m\therefore n_1=\dfrac{1}{2l}\sqrt{\dfrac{w_1}{m}} and n2=12lw2mn_2=\dfrac{1}{2l}\sqrt{\dfrac{w_2}{m}} ∴n1n2=w1w2=98=1.06\therefore \dfrac{n_1}{n_2}=\sqrt{\dfrac{w_1}{w_2}}=\sqrt{\dfrac{9}{8}}=1.06 ∴n1>n2....(i)\therefore n_1>n_2\qquad....(i) Beat frequency n1−n2=4n_1-n_2=4 ∴n2=n1−4....(ii)\therefore n_2=n_1-4\qquad....(ii) From (i) and (ii), ∴n1n1−4=1.06\therefore \dfrac{n_1}{n_1-4}=1.06 ∴n1=4×1.060.06\therefore n_1=\dfrac{4\times1.06}{0.06} =70.67=70.67 Hz Ans: The frequency of the fork is 70.67 Hz.

  3. Given: l = 36 cm = 0.36 m, n = 280 Hz, T = 24.5 N To find: Linear density (m) Formula: n=12lTmn=\dfrac{1}{2l}\sqrt{\dfrac{T}{m}} Calculation: Using formula, m=T4l2n2=24.54×0.362×2802m=\dfrac{T}{4l^2n^2}=\dfrac{24.5}{4\times0.36^2\times280^2} = antilog{log(24.5) − [log(4) + 2 log(0.36) + 2 log(280)]} = antilog{1.3892 − [0.6021 + 2(1‾\overline1.5563) + 2(2.4472)]} = antilog{1.3892 − (0.6021 + 1‾\overline1.1126 + 4.8944)} = antilog{1.3892 − 4.6091} = antilog{4‾\overline4.7801} =6.027×10−4=6.027\times10^{-4} kg/m Ans: Linear density of material of wire is 6.027×10−46.027\times10^{-4} kg/m.

  4. Given: When wire is stretched by a weight hanging vertically, n1=125n_1=125 Hz, Let frequency of wire when the weight is immersed in water producing 8 beats per second =n2=n_2 ∴n2=n1−8=125−8=117\therefore n_2=n_1-8=125-8=117 Hz To find: Specific gravity (ρ) Formula: n1n2=ρρ−1\dfrac{n_1}{n_2}=\sqrt{\dfrac{\rho}{\rho-1}} Calculation: From formula, 125117=ρρ−1\dfrac{125}{117}=\sqrt{\dfrac{\rho}{\rho-1}} 1.068=ρρ−11.068=\sqrt{\dfrac{\rho}{\rho-1}} ∴1.141=ρρ−1\therefore 1.141=\dfrac{\rho}{\rho-1} ∴ρ=1.141(ρ−1)\therefore \rho=1.141(\rho-1) ∴ρ=1.141ρ−1.141\therefore \rho=1.141\rho-1.141 1.141=1.141ρ−ρ1.141=1.141\rho-\rho ∴ρ=1.1410.141=\therefore \rho=\dfrac{1.141}{0.141}= 8.09 Ans: The specific gravity of material of the weight is 8.09.

  5. Given: l2=1.05l1⇒l2l1=1.05l_2=1.05l_1\Rightarrow\dfrac{l_2}{l_1}=1.05, N = 10 s−1^{-1} To find: Frequency (n1)(n_1) Formula: For two wires of same material, n1l1=n2l2n_1l_1=n_2l_2 Calculation: In first case, n1=12l1Tmn_1=\dfrac{1}{2l_1}\sqrt{\dfrac{T}{m}} In second case, n2=12l2Tmn_2=\dfrac{1}{2l_2}\sqrt{\dfrac{T}{m}} Since, l2>l1l_2>l_1, then n1>n2n_1>n_2 ∴n1−n2=10\therefore n_1-n_2=10 ∴n2=n1−10\therefore n_2=n_1-10 Given that, l2=1.05l1l_2=1.05l_1 ∴n1l1=(n1−10)(1.05l1)\therefore n_1l_1=(n_1-10)(1.05l_1) ∴n1=(n1−10)(1.05)\therefore n_1=(n_1-10)(1.05) 1.05n1−n1=10.51.05n_1-n_1=10.5 ∴0.05n1=10.5\therefore 0.05n_1=10.5 ∴n1=10.50.05=\therefore n_1=\dfrac{10.5}{0.05}= 210 Hz Ans: The frequency of the fork is 210 Hz.

  6. Given: l = 1 m, M = 2 g = 2×10−32\times10^{-3} kg, n = 300 Hz ∴m=Ml=2×10−3\therefore m=\dfrac{M}{l}=2\times10^{-3} kg m−1^{-1} To find: Tension in the wire (T) Formula: n=12lTmn=\dfrac{1}{2l}\sqrt{\dfrac{T}{m}} Calculation: From formula, n2=14l2×Tmn^2=\dfrac{1}{4l^2}\times\dfrac{T}{m} ∴T=m×n2×4l2\therefore T=m\times n^2\times4l^2 =2×10−3×(300)2×4×(1)2==2\times10^{-3}\times(300)^2\times4\times(1)^2= 720 N Ans: The tension in the sonometer wire is 720 N.