
The speed of a wave of time period T and propagation constant K is
A. $\dfrac{{{{2\pi }}}}{{{{TK}}}}$
B. $\dfrac{{{{TK}}}}{{{{2\pi }}}}$
C. $\dfrac{{{1}}}{{{{TK}}}}$
D. $\dfrac{{{T}}}{{{K}}}$
Answer
506.1k+ views
Hint: In order to find the correct relation we should know the relation between propagation constant K and wavelength $\lambda $. Wave behavior waves display several basic phenomena. In reflection, a wave encounters an obstacle and is reflected back. In refraction, a wave bends when it enters a medium through which it has a different speed.
Formula used:
${{\upsilon = }}\dfrac{{{\lambda }}}{{{T}}}$
$\upsilon = $ velocity of wave
${{T}}$ $ = $ time period
$\lambda = $ wavelength
(ii) $K = \dfrac{{2\pi }}{\lambda }$ $\because $ K is constant, $\lambda $ is wavelength.
Complete step by step answer:
Velocity of wave is ${{\lambda = }}\dfrac{{{\lambda }}}{{{T}}}$
Now,
Propagation constant K and wavelength are related as
${{K}}{{ = }}\,\dfrac{{{{2\pi }}}}{{{\lambda }}}{{or}}{{\lambda }}{{ = }}\dfrac{{{{2\pi }}}}{{{K}}}$
${{Q}}\;{\text{velocity}}\;{{V}}{{ = }}\dfrac{{{\lambda }}}{{{T}}}{{ = }}\dfrac{{{{2\pi }}}}{{{{KT}}}}$.
Hence, the correct answer is option (A).
Additional information:
In diffraction, waves bend when they pass around small obstacles and spread out when they pass through small openings. In interference, when two waves meet, they can interfere constructively, creating a wave with larger amplitude than the original wave with larger amplitude or destructively creating a wave with smaller (or even zero) amplitude.
The production wave takes place when a vibrating source disturbs the first particle of a periodical nature. This certainly creates a wave pattern that begins to travel from particle to particle along with a medium. Furthermore the frequency at the vibration of a particle occurs and is equal to the frequency of sound vibration.
${\text{Speed}}{{ = }}\dfrac{{{\text{Wavelength}}}}{{{\text{Period}}}}{{ = }}\dfrac{{{\lambda }}}{{{T}}}$
Or
${\text{Speed}}{{ = }}{\text{wavelength} \times \text{frequency }} \Rightarrow {{ S = }}\lambda \times \upsilon $
Or
${{V}}{{ = }}\upsilon {{ \times \lambda }}$
$\upsilon = $ frequency of wave.
$\lambda = $ wavelength of wave.
$V = $ speed of wave.
Note: By knowing the relation between propagation constant K & wavelength $\lambda $ we can find the correct option i.e. ${{K}}{{ = }}\dfrac{{{{2\pi }}}}{{{\lambda }}}$ and ${{V}}{{ = }}\dfrac{{{\lambda }}}{{{T}}}$. The question could also be solved by dimensional analysis or just be check the unit of each option.
Formula used:
${{\upsilon = }}\dfrac{{{\lambda }}}{{{T}}}$
$\upsilon = $ velocity of wave
${{T}}$ $ = $ time period
$\lambda = $ wavelength
(ii) $K = \dfrac{{2\pi }}{\lambda }$ $\because $ K is constant, $\lambda $ is wavelength.
Complete step by step answer:
Velocity of wave is ${{\lambda = }}\dfrac{{{\lambda }}}{{{T}}}$
Now,
Propagation constant K and wavelength are related as
${{K}}{{ = }}\,\dfrac{{{{2\pi }}}}{{{\lambda }}}{{or}}{{\lambda }}{{ = }}\dfrac{{{{2\pi }}}}{{{K}}}$
${{Q}}\;{\text{velocity}}\;{{V}}{{ = }}\dfrac{{{\lambda }}}{{{T}}}{{ = }}\dfrac{{{{2\pi }}}}{{{{KT}}}}$.
Hence, the correct answer is option (A).
Additional information:
In diffraction, waves bend when they pass around small obstacles and spread out when they pass through small openings. In interference, when two waves meet, they can interfere constructively, creating a wave with larger amplitude than the original wave with larger amplitude or destructively creating a wave with smaller (or even zero) amplitude.
The production wave takes place when a vibrating source disturbs the first particle of a periodical nature. This certainly creates a wave pattern that begins to travel from particle to particle along with a medium. Furthermore the frequency at the vibration of a particle occurs and is equal to the frequency of sound vibration.
${\text{Speed}}{{ = }}\dfrac{{{\text{Wavelength}}}}{{{\text{Period}}}}{{ = }}\dfrac{{{\lambda }}}{{{T}}}$
Or
${\text{Speed}}{{ = }}{\text{wavelength} \times \text{frequency }} \Rightarrow {{ S = }}\lambda \times \upsilon $
Or
${{V}}{{ = }}\upsilon {{ \times \lambda }}$
$\upsilon = $ frequency of wave.
$\lambda = $ wavelength of wave.
$V = $ speed of wave.
Note: By knowing the relation between propagation constant K & wavelength $\lambda $ we can find the correct option i.e. ${{K}}{{ = }}\dfrac{{{{2\pi }}}}{{{\lambda }}}$ and ${{V}}{{ = }}\dfrac{{{\lambda }}}{{{T}}}$. The question could also be solved by dimensional analysis or just be check the unit of each option.
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