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Question

Physics Question on Waves

A vibrating tuning fork of frequency υ\upsilon is placed near the open end of a long cylindrical tube. The tube has a side opening and is also fitted with a movable reflecting piston. As the piston is moved through 8.75cm8.75 \,cm, the intensity of sound changes from a maximum to minimum. If the speed of sound is 350ms1350 \,m \,s^{-1}, then υ\upsilon is

A

500Hz500\, Hz

B

1000Hz1000 \, Hz

C

2000Hz2000 \, Hz

D

4000Hz4000 \, Hz

Answer

1000Hz1000 \, Hz

Explanation

Solution

Additional path difference introduced when the piston is moved through 8.75cm=2×8.75=17.50cm8.75\, cm =2 \times 8.75=17.50\,cm Since, the intensity changes from a maximum to a minimum additional path = λ/2\lambda/ 2 (λ2)=17.50\therefore \left(\lambda 2\right)=17.50 or λ=35.0cm\lambda=35.0\, cm Velocity, v=350ms1=35000cms1 v=350\,m\,s^{-1}=35000\,cm \,s^{-1} \therefore Frequency, υ=vλ=3500035=1000Hz \upsilon=\frac{v}{\lambda}=\frac{35000}{35}=1000\,Hz