Sunday, April 01, 2007

MCQ on Vibration of Strings & Sonometer

In continuation of the post dated 18th March 2007, let us discuss two more questions involving transverse waves in stretched strings. The following MCQ appeared in Kerala Engineering Entrance Examination 2006 question paper:

Two stretched strings of same material are vibrating under same tension in fundamental mode. The ratio of their frequencies is 1:2 and ratio of the lengths of the vibrating segments is 1:4. Then the ratio of the radii of the strings is
(a) 2:1 (b) 4:1 (c) 3:2 (d) 8:1 (e) 4:5
You know that the frequency (n) of vibration of a stretched string in the fundamental mode is given by
n = (1/2L)√(T/m) = (1/2L)√(T/πr2ρ) where L is the length of the vibrating segment of the wire, T is the tension and ‘m’ is the linear density (mass per unit length) which is πr2ρ where ‘r’ is the radius and ρ is the density of the material of the wire.
Therefore, if n1 and n2 are the frequencies of the two wires having lengths L1 and L2 and radii r1 and r2 respectively, we have
n1/n2 = [(1/2L1)√(T/πr12 ρ)] / [(1/2L2)√(T/πr22ρ) = L2r2/L1r1. [Note that the strings have the same density since they are of the same material].
Since n1/n2 = ½ and L1/L2 = ¼ as given in the question,
½ = 4r2/r1 from which r1/r2 = 8 [Option (d)].

Now, consider the following question:
A sonometer wire is kept stretched by suspending a 60 kg mass from the free end of the wire. The suspended mass has a volume of 0.01 m3. The fundamental frequency of the wire in this condition is 300 Hz. If the suspended mass is completely immersed in water, the fundamental frequency of the wire will be approximately
(a) 285 Hz (b) 273 Hz (c) 300 Hz (d) 330 Hz (e) 355 Hz
The frequency will decrease since the tension is decreased (because of the decreased weight of the mass in water).
When the mass is in air, the tension,T1 = mg = 60×g where ‘g’ is the acceleration due to gravity.
When the mass is in water, tension, T2 = mg – up thrust = 60×g – 0.01×1000×g = 50×g. [Note that the up thrust is the weight of 0.01 m3 of water displaced by the suspended mass].
Since the frequency is given by n = (1/2L)√(T/m) with usual notations, n α √T.
If n1 and n2 are the frequencies in the two cases, n1/n2 = √(T1/T2) or,
300/n2 = √(60/50) = √1.2 = 1.1 (nearly) so that n2 = 273 (nearly).

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