A tuning fork of frequency 200 Hz is in unison with a sonometer wire. The number of beats heard per second when the tension is increased by 1 % is

Question

A tuning fork of frequency 200 Hz is in unison with a sonometer wire. The number of beats heard per second when the tension is increased by 1 % is (A) 1 (B) 2 (C) 4 (D) 1/2

Tardigrade Admin · Accepted Answer

(n2/n1)=√(101/100)=(1+(1/100))1 / 2=1+(1/200) (From Binomial Expansion) n2=n1+(n1/200) Number of beats s -1=n2-n1=(n1/200)=(200/200)=1

Q. A tuning fork of frequency $200 Hz$ is in unison with a sonometer wire. The number of beats heard per second when the tension is increased by $1%$ is

1

2

4

1/2

$\frac{n _{2}}{n _{1}} = \frac{101}{100} = (1 + \frac{1}{100})^{1/2} = 1 + \frac{1}{200}$
(From Binomial Expansion)
$n_{2} = n_{1} + \frac{n _{1}}{200}$
Number of beats $s^{- 1} = n_{2} - n_{1} = \frac{n _{1}}{200} = \frac{200}{200} = 1$

Q. A tuning fork of frequency 200Hz is in unison with a sonometer wire. The number of beats heard per second when the tension is increased by 1% is

1

2

4

1/2

n1​n2​​=100101​​=(1+1001​)1/2=1+2001​ (From Binomial Expansion) n2​=n1​+200n1​​ Number of beats s−1=n2​−n1​=200n1​​=200200​=1

Q. A tuning fork of frequency $200 Hz$ is in unison with a sonometer wire. The number of beats heard per second when the tension is increased by $1%$ is

$\frac{n _{2}}{n _{1}} = \frac{101}{100} = (1 + \frac{1}{100})^{1/2} = 1 + \frac{1}{200}$
(From Binomial Expansion)
$n_{2} = n_{1} + \frac{n _{1}}{200}$
Number of beats $s^{- 1} = n_{2} - n_{1} = \frac{n _{1}}{200} = \frac{200}{200} = 1$