Question

The wavelengths of two waves are 90 cm and 91 cm. If the room temperature is 45${}^o$C then the number of beats produced per second by these waves will be (velocity of sound at 0${}^o$C=333 m/s)

• A. 4.4
• B. 10
• C. 8
• D. 12

Answer 1

The correct option is A 4.4

Velocity of sound in air $v\propto \sqrt{T}$
$⟹$ $\frac{v_T}{v_o}=\sqrt{\frac{T_2}{T_1}}$
Given: $v_o=333m/s$, $T_1=273K$ and $T_2=273+45=318K$
$\therefore$ $\frac{v_T}{333}=\sqrt{\frac{318}{273}}$
$⟹v_T=359.27m/s=35927cm/s$
Also, given ${\lambda }_1=90cm$ and ${\lambda }_2=91cm$
Beat frequency $b=|{\nu }_1-{\nu }_2|=v_T(\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2})$
$\therefore$ $b=\left(35927\right)(\frac{1}{90}-\frac{1}{91})=4.4Hz$