A free response practice question involving waves and electromagnetism was given to you in the post dated 30^{th} October 2009. As promised, I give below a model answer along with the question:

*steel*wire of linear density

*m*, standard weights, a small electromagnet for vibrating the wire, an AC source of variable output voltage but of fixed frequency (which is to be measured), connecting wires and metre scale. Now answer the following questions:

**(a)** Show in a diagram how you will arrange the given items for the measurement of the frequency of the given AC source.

**(b)** You are given that the frequency of vibration of the sonometer wire segment of length ℓ in the s^{th} mode is (s/2ℓ)√(*T*/*m*) where *T* is the tension.

(i) Now explain briefly how you will proceed to measure the frequency of the alternating current making use of the vibration of the sonometer wire segment in the *fundamental *mode.

(ii) Write down the expression for the frequency of the wire.

(ii) How is the frequency of AC related to the frequency of vibration of the sonometer wire? Justify your answer.

**(c)** Your teacher now wants you to use a *brass* wire of linear density *μ *instead of the steel wire in the sonometer. Instead of the electromagnet she gives you a sufficiently strong permanent bar magnet.

Show in a diagram how you will arrange the given items for the measurement of the frequency of the given AC source.

**(d)** (i) Explain briefly how you will proceed to measure the frequency of the alternating current in this case.

(ii) How is the frequency of AC related to the frequency of vibration of the sonometer wire in this case? Justify your answer.

**(a)** The arrangement for measuring the frequency is shown in the following figure in which the electromagnet is shown connected to the AC source:

**(b)** (i) By suspending a suitable mass *M*, the steel wire in the sonometer is kept under tension *T *(= *Mg*). The electromagnet energized by the AC source is held above the steel wire segment AB in the sonometer. The wire oscillates because of the periodic attraction by the electromagnet. The length of the wire segment AB is adjusted by moving the knife edge A or B of the sonometer so that the wire segment is at resonance. This condition is *minimum* length ℓ of the wire segment for attaining the resonance condition is measured using a metre scale.

(ii) The frequency of vibration of the sonometer wire segment of length ℓ in the s^{th} mode is (s/2ℓ)√(*T*/*m*) where *T* is the tension and *m* is the linear density of the wire. In the fundamental mode, s = 1. Therefore, the frequency of vibration of the sonometer wire segment = (1/2ℓ)√(*T*/*m*).

(iii) Since the steel wire is attracted towards the electromagnet during the negative as well as positive *twice* the frequency of the AC.

Therefore, frequency of the AC = (1/4ℓ)√(*T*/*m*).

**(c)** For the measurement of the frequency of AC using the vibration of the brass wire, the alternating current is passed through the brass wire (in the sonometer) by connecting it in series with the AC source. The bar magnet is arranged near the wire so that the axis of the magnet is horizontal and at right angles to the wire. The arrangement is shown in the following figure

**(d)**

(i) The current carrying wire of the sonometer is located in the magnetic field produced by the bar magnet and hence it will experience magnetic forces. Since the current is alternating, the force is periodic and the wire will vibrate. Keeping the wire under a suitable tension *T*_{1 }(= *M*_{1}g), the length of the wire segment AB is adjusted by moving the knife edge A or B of the sonometer so that resonance is attained. The *minimum* length of the wire segment ℓ_{1} (corresponding to the fundamental mode of vibration) is measured. The frequency of vibration of the wire is (1/2ℓ_{1} )√(*T*_{1}/*μ*). ** **

(ii) The current carrying brass wire is located in the magnetic field of the bar magnet and hence feels a magnetic force. The wire is pulled up during one half cycle of AC and pushed down during the next half cycle since the direction of the current in the wire changes. This means that the frequency of vibration of the wire is the *same* as that of the AC passing through the wire. Therefore, frequency of AC = (1/2ℓ_{1} )√(*T*_{1}/*μ*).

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