First slide

Standing waves

Question

A uniform rod AB of length 0 . 4 m and mass 1 .2 kg is suspended by two identical wires from a ceiling as shown in fig. (3). Where should a mass of 4.8 kg be
suspended from the rod so that a tuning fork of frequency 256 Hz is in resonance with the first wire at Awhile its first overtone resonates with second wire at B?

Moderate

A

5 cm
B

10 cm
C

15 cm
D

20 cm

Solution

$Here n_{1} = \frac{1}{2 l} \sqrt{(\frac{T_{1}}{m})} = 256 and n_{2} = \frac{1}{2 l} \sqrt{(\frac{T_{2}}{m}) = 2 \times 256} From these equations, we have T_{1} = 4 T_{2} Let the mass be suspended at a distance x ftom A T_{1} + T_{2} = 4 \cdot 8 + 1 \cdot 2 = 6 Taking moments aboutA for equilibrium 1 \cdot 2 \times 0 \cdot 2 + 4 \cdot 8 \times x = T_{2} \times 0 \cdot 4 From eqs. (1) and (2), T_{2} = 1 .2 1 \cdot 2 \times 0 \cdot 2 + 4 \cdot 8 \times x = 1 \cdot 2 \times 0 \cdot 4 x = 5 cm$

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