
Vibrating tuning fork of frequency $n$ is placed near the open end of a long cylindrical tube. The tube has a side opening and is fitted with a movable reflecting piston. As the piston is moved through $8.75cm$, the intensity of sound changes from a maximum to minimum. If the speed of sound is $350m/s$ . Then $n$ is

A. $500\,Hz$
B. $1000\,Hz$
C. $2000\,Hz$
D. $4000\,Hz$
Answer
164.1k+ views
Hint: This problem is based on sound waves where we have to find $n$ i.e., frequency. We know that to find the frequency we must have the value of wavelength. As the intensity changes from maximum to minimum, the path difference will be $2l$. Use mathematical relation $f = \dfrac{v}{\lambda }$ to find the frequency of a sound wave.
Formula used:
$\text{Path Difference} = \dfrac{{(n + 1)\lambda }}{2} - \dfrac{{(n)\lambda }}{2}$
$\Rightarrow \text{Path Difference}= \dfrac{\lambda }{2}$
Here, $\lambda$ is the wavelength of sound.
$f(\text{frequency}) = \dfrac{{v(\text{velocity})}}{{\lambda (\text{wavelength})}}$
Complete step by step solution:
As the piston is moved through $l = 8.75\,cm$ (given), therefore path difference must be$2l = 2 \times 8.75 = 17.5\,cm$ … (1)
Also, the intensity changes from maximum to minimum (given)
$\text{Path Difference} = \dfrac{{(n + 1)\lambda }}{2} - \dfrac{{(n)\lambda }}{2}$
$\Rightarrow \text{Path Difference}= \dfrac{\lambda }{2}$
i.e., From equation (1), we get
$\dfrac{\lambda }{2} = 17.5cm$
$ \Rightarrow \lambda = 35cm = 0.35m$
We know that,
$f(\text{frequency}) = \dfrac{{v(\text{velocity})}}{{\lambda (\text{wavelength})}}$
In this case,
$ \Rightarrow n = \dfrac{v}{\lambda } = \dfrac{{350}}{{0.35}}$
$ \therefore n = 1000\,Hz$
Thus, the frequency of a sound wave produced by a tuning fork due to a change in intensity of sound from a maximum to minimum is $1000\,Hz$.
Hence, the correct option is B.
Note: Since this is a numerical problem on sound waves hence, it is essential that phase difference must be calculated first as it will help us in finding the wavelength then the given parameters must be used very carefully to give an accurate solution. While writing an answer, always remember to put the units after results.
Formula used:
$\text{Path Difference} = \dfrac{{(n + 1)\lambda }}{2} - \dfrac{{(n)\lambda }}{2}$
$\Rightarrow \text{Path Difference}= \dfrac{\lambda }{2}$
Here, $\lambda$ is the wavelength of sound.
$f(\text{frequency}) = \dfrac{{v(\text{velocity})}}{{\lambda (\text{wavelength})}}$
Complete step by step solution:
As the piston is moved through $l = 8.75\,cm$ (given), therefore path difference must be$2l = 2 \times 8.75 = 17.5\,cm$ … (1)
Also, the intensity changes from maximum to minimum (given)
$\text{Path Difference} = \dfrac{{(n + 1)\lambda }}{2} - \dfrac{{(n)\lambda }}{2}$
$\Rightarrow \text{Path Difference}= \dfrac{\lambda }{2}$
i.e., From equation (1), we get
$\dfrac{\lambda }{2} = 17.5cm$
$ \Rightarrow \lambda = 35cm = 0.35m$
We know that,
$f(\text{frequency}) = \dfrac{{v(\text{velocity})}}{{\lambda (\text{wavelength})}}$
In this case,
$ \Rightarrow n = \dfrac{v}{\lambda } = \dfrac{{350}}{{0.35}}$
$ \therefore n = 1000\,Hz$
Thus, the frequency of a sound wave produced by a tuning fork due to a change in intensity of sound from a maximum to minimum is $1000\,Hz$.
Hence, the correct option is B.
Note: Since this is a numerical problem on sound waves hence, it is essential that phase difference must be calculated first as it will help us in finding the wavelength then the given parameters must be used very carefully to give an accurate solution. While writing an answer, always remember to put the units after results.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement
