
A person listening to a tone of $500{\text{ Hz}}$ sitting at a distance $450{\text{ m}}$ from the source of the sound. What is the time interval between the successive compressions from the source?
$\left( a \right){\text{ 5 ms}}$
$\left( b \right){\text{ 1 ms}}$
$\left( c \right){\text{ 2 ms}}$
$\left( d \right){\text{ 2 s}}$
Answer
220.5k+ views
Hint Frequency is the number of cycles per second meant as Hertz. The period is seconds per cycle. Converse since, supposing that the frequency is high, at that point the period is low. So by using the above statement we will be able to find it.
Formula used:
Time period,
$T = \dfrac{1}{f}$
Here,
$T$, will be the time period
$f$, will be the frequency.
Complete Step By Step Solution Frequency and period are contrary related amounts. So, the time span between progressive compressions is equivalent to the time-frame of the wave. Furthermore, this time span is proportional to the recurrence of the wave.
Therefore, by using the formula for the time period
We get
$T = \dfrac{1}{f}$
Now substitute the values, we get
$ \Rightarrow T = \dfrac{1}{{500}}s$
Now on solving the above equation, we get
$ \Rightarrow T = 0.002{\text{ s}}$
Therefore, in milliseconds it will be
$ \Rightarrow T = 2{\text{ ms}}$
The above time period will be required for the source having successive compression.
Hence the option $\left( c \right)$ will be correct.
Additional information At the point when a function happens consistently, at that point, we state that the function is occasional and allude to the ideal opportunity for the function to rehash itself as the period. Period - as expected - is estimated in a flash, hours, days, or years. For example, the time of the pivot of Earth on its hub is twenty-four hours.
Note There is the relationship that the time period is the average distance between bodies or particles or photons divided by their average velocity. The frequency is the reciprocal of the period. The average is measured over some time if it varies, or it is necessary to do so to measure a minimum quantity.
Formula used:
Time period,
$T = \dfrac{1}{f}$
Here,
$T$, will be the time period
$f$, will be the frequency.
Complete Step By Step Solution Frequency and period are contrary related amounts. So, the time span between progressive compressions is equivalent to the time-frame of the wave. Furthermore, this time span is proportional to the recurrence of the wave.
Therefore, by using the formula for the time period
We get
$T = \dfrac{1}{f}$
Now substitute the values, we get
$ \Rightarrow T = \dfrac{1}{{500}}s$
Now on solving the above equation, we get
$ \Rightarrow T = 0.002{\text{ s}}$
Therefore, in milliseconds it will be
$ \Rightarrow T = 2{\text{ ms}}$
The above time period will be required for the source having successive compression.
Hence the option $\left( c \right)$ will be correct.
Additional information At the point when a function happens consistently, at that point, we state that the function is occasional and allude to the ideal opportunity for the function to rehash itself as the period. Period - as expected - is estimated in a flash, hours, days, or years. For example, the time of the pivot of Earth on its hub is twenty-four hours.
Note There is the relationship that the time period is the average distance between bodies or particles or photons divided by their average velocity. The frequency is the reciprocal of the period. The average is measured over some time if it varies, or it is necessary to do so to measure a minimum quantity.
Recently Updated Pages
Two discs which are rotating about their respective class 11 physics JEE_Main

A ladder rests against a frictionless vertical wall class 11 physics JEE_Main

Two simple pendulums of lengths 1 m and 16 m respectively class 11 physics JEE_Main

The slopes of isothermal and adiabatic curves are related class 11 physics JEE_Main

A trolly falling freely on an inclined plane as shown class 11 physics JEE_Main

The masses M1 and M2M2 M1 are released from rest Using class 11 physics JEE_Main

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Understanding Uniform Acceleration in Physics

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

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

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

NCERT Solutions For Class 11 Physics Chapter 8 Mechanical Properties Of Solids

Motion in a Straight Line Class 11 Physics Chapter 2 CBSE Notes - 2025-26

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

