A tuning fork of frequency 200 Hz is in unison with a sonometer wire. Tension in the wire of sonometer is increased by 1 % without any change in its length. Find the number of beats heard in 9 s .

Question

A tuning fork of frequency 200 Hz is in unison with a sonometer wire. Tension in the wire of sonometer is increased by 1 % without any change in its length. Find the number of beats heard in 9 s . (A) 9 (B) 3 (C) 6 (D) 12

Tardigrade Admin · Accepted Answer

f=(1/2 π )√(T/μ ) (Δ f/f)=(1/2)(âT/T) Δ f=(1/2)× (1/100)× 200 Δ f=1 Beat frequency = 1 Number of beats in 9 second =9× 1=9

Q. A tuning fork of frequency $200 Hz$ is in unison with a sonometer wire. Tension in the wire of sonometer is increased by $1%$ without any change in its length. Find the number of beats heard in $9 s$ .

$9$

$3$

$6$

$12$

$f = \frac{1}{2 π} \frac{T}{μ}$
$\frac{Δ f}{f} = \frac{1}{2} \frac{∆ T}{T}$
$Δ f = \frac{1}{2} \times \frac{1}{100} \times 200$
$Δ f = 1$
Beat frequency = 1
Number of beats in 9 second $= 9 \times 1 = 9$

Q. A tuning fork of frequency 200Hz is in unison with a sonometer wire. Tension in the wire of sonometer is increased by 1% without any change in its length. Find the number of beats heard in 9s .

9

3

6

12

f=2π1​μT​​ fΔf​=21​T∆T​ Δf=21​×1001​×200 Δf=1 Beat frequency = 1 Number of beats in 9 second =9×1=9

Q. A tuning fork of frequency $200 Hz$ is in unison with a sonometer wire. Tension in the wire of sonometer is increased by $1%$ without any change in its length. Find the number of beats heard in $9 s$ .

$9$

$3$

$6$

$12$

$f = \frac{1}{2 π} \frac{T}{μ}$
$\frac{Δ f}{f} = \frac{1}{2} \frac{∆ T}{T}$
$Δ f = \frac{1}{2} \times \frac{1}{100} \times 200$
$Δ f = 1$
Beat frequency = 1
Number of beats in 9 second $= 9 \times 1 = 9$