Page 28 - 11-phy-8 Waves

8. Waves eLearn.Punjab

Example 8.3: A steel wire hangs vertically from a fixed point, Do You Know?

supporting a weight of 80 N at its lower end. The diameter of the
wire is 0.50 mm and its length from the fixed point to the weight
is 1.5 m. C alculate the fundamental frequency emitted by the
wire when it is plucked?
(Density of steel wire = 7.8 x 103 kgm -3)

Solution :
Volume of wire = Length× Area of cross section
Mass = Volume × Density

therefore A standing-wave pattern is formed when
Mass of the wire = Length× Area of cross section× Density the length of the string is an integral
So mass per unit lengthm is given by multiple of half wave-length; otherwise
no standing wave is formed.
m = Density × Area of cross section
For Your Information
Diameter of the wire = D = 0.50mm = 0.5 ×10-3
Radius of the wire = r = D = 0.25 ×10-3m In an organ pipe, the primary driving
mechanism is wavering, sheet
2 like jet of air from flute-slit, which
interacts with the upper lip and the
Area of cross section of wire = pr2 = 3.14 × (0.25 ×10-3m)2 air column in the pipe to maintain a
F =w steady oscillation.

therefore v: 1.1
m = 7.8 ×103kgm-3 × 3.14 × (0.25 ×10-3m)2
m = 1.53 ×10-3kgm-1
Weight = 80N = 80 kgms-2

Using the equation (8.17), we get

f1 = 1 F
2l m

f1 = 2 1 m 80 kgms-2 = 76 s-1
×1.5 1.53 ×10-3kgm-1

or f1 = 76 Hz.

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