Question

Two wires are fixed on a sonometer. Their tensions are in the
ratio 8:1, their lengths are in the ratio 36:35, the diameters
are in the ratio 4:1 and densities are in the ratio 1:2. If the
note of the higher pitch has a frequency 360 s', the frequency
of beats produced is
(b) 10 s-
(d) 20 s-
(a) 55
(c) 15 s-​

Answer 1

Answer:

your ans is b option hope you like it please follow

Answer 2

The frequency of of beats produced is:

(b) 10 s⁻¹

This can be calculated as follows:

• We know that, Frequency (η) =  $\frac{1}{2l} \sqrt{\frac{T}{m} }$

                                                        = $\frac{1}{2l} \sqrt{\frac{T}{\pi r^{2} p } }$

                                                        = $\frac{1}{2lr} \sqrt{\frac{T}{\pi p } }$

• Also, (η₁/η₂) = (l₂.r₂)/(l₁r₁) $\sqrt{\frac{T_{1} }{T_{2} } }$  $\sqrt{\frac{p_{1} }{p_{2} } }$

                            =  (35 x 1/36 x 6) $\sqrt{\frac{8 }{1 } }$   $\sqrt{\frac{2 }{1 } }$

                            = (35/36)

• As per the question, higher frequency (η₂) is 360 s⁻¹.

• Therefore, η₁ = (35/36) x 360

                              =350

• Therefore, beat frequency = 360 - 350

                                                     = 10s⁻¹