364305
A body is executing simple harmonic motion with an angular frequency of \(2\,rad{s^{ - 1}}.\) The velocity of the body at \(20\;mm\) displacement, when the amplitude of the motion is \(60\;mm\) is
364306
The velocity of a particle in simple harmonic motion at displacement \(y\) from mean position is
1 \(\omega \sqrt{a^{2}-y^{2}}\)
2 \(\omega \sqrt{a^{2}+y^{2}}\)
3 \(\omega^{2} \sqrt{a^{2}-y^{2}}\)
4 \(\omega y\)
Explanation:
Conceptual Question
PHXI14:OSCILLATIONS
364307
Graph between velocity and displacement of a particle, executing S.H.M. is
1 A parabola
2 A straight line
3 An ellipse
4 A hyperbola
Explanation:
In simple harmonic motion \(y=a \sin \omega t\) and \(v=\omega \cos \omega t\) from this we have \(\dfrac{y^{2}}{a^{2}}+\dfrac{v^{2}}{a^{2} \omega^{2}}=1\), which is equation of ellipse.
PHXI14:OSCILLATIONS
364308
If ' \(x\) ', ' \(v\) 's, and a denote the displacement velocity and acceleration of a particle respectively executing SHM of preiodic time \(t\), then which one of the following does not change with time?
1 \(\dfrac{a T}{x}\)
2 \(a t+2 \pi v\)
3 \(\dfrac{a T}{v}\)
4 \(a t+4 \pi^{2} v^{2}\)
Explanation:
The dimensions of given variables of SHM are as Displacement \([x]=\left[M^{0} L T^{0}\right]\) Velocity \([v]=\left[M^{0} L T^{-1}\right]\) Acceleration \([a]=\left[M^{0} L T^{-2}\right]\) and time period, \([T]=\left[M^{0} L^{0} T\right]\) Now, checking each option for these values For option(1) \(\dfrac{[a][T]}{[x]}=\dfrac{\left[M^{0} L T^{-2}\right]\left[M^{0} L^{0} T\right]}{\left[M^{0} L T^{0}\right]}=\left[M^{0} L^{0} T^{-1}\right]\) As it depends on time, so change with it. For option(2) \([a][T]+2 \pi[v]=\) \(\left[M^{0} L T^{-2}\right]\left[M^{0} L^{0} T\right]+\left[M^{0} L^{0} T^{-1}\right]\) \(=\left[M^{0} L T^{-1}\right]\) it is also dependent on time and hence changes with it For option (3) \(\begin{aligned}\dfrac{[a][T]}{[v]} & =\dfrac{\left[M^{0} L T^{-2}\right]\left[M^{0} L^{0} T\right]}{\left[M^{0} L T^{-1}\right]} \\& =\left[M^{0} L^{0} T^{0}\right]\end{aligned}\) As it is a constant having no dimension, so it does not change in time For option (4), \(\begin{aligned}& {[a][t]+4 \pi^{2}[d]^{2}=} \\& {\left[M^{0} L T^{-2}\right]\left[M^{0} L^{0} T\right]+\left[M^{0} L T^{-1}\right]^{2}} \\& =\left[L T^{-1}\right]+\left[L^{2} T^{-2}\right]\end{aligned}\) As, the term is dependent on time, so changes with it Also, it is dimensionally incorrect Hence option (3) is correct
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PHXI14:OSCILLATIONS
364305
A body is executing simple harmonic motion with an angular frequency of \(2\,rad{s^{ - 1}}.\) The velocity of the body at \(20\;mm\) displacement, when the amplitude of the motion is \(60\;mm\) is
364306
The velocity of a particle in simple harmonic motion at displacement \(y\) from mean position is
1 \(\omega \sqrt{a^{2}-y^{2}}\)
2 \(\omega \sqrt{a^{2}+y^{2}}\)
3 \(\omega^{2} \sqrt{a^{2}-y^{2}}\)
4 \(\omega y\)
Explanation:
Conceptual Question
PHXI14:OSCILLATIONS
364307
Graph between velocity and displacement of a particle, executing S.H.M. is
1 A parabola
2 A straight line
3 An ellipse
4 A hyperbola
Explanation:
In simple harmonic motion \(y=a \sin \omega t\) and \(v=\omega \cos \omega t\) from this we have \(\dfrac{y^{2}}{a^{2}}+\dfrac{v^{2}}{a^{2} \omega^{2}}=1\), which is equation of ellipse.
PHXI14:OSCILLATIONS
364308
If ' \(x\) ', ' \(v\) 's, and a denote the displacement velocity and acceleration of a particle respectively executing SHM of preiodic time \(t\), then which one of the following does not change with time?
1 \(\dfrac{a T}{x}\)
2 \(a t+2 \pi v\)
3 \(\dfrac{a T}{v}\)
4 \(a t+4 \pi^{2} v^{2}\)
Explanation:
The dimensions of given variables of SHM are as Displacement \([x]=\left[M^{0} L T^{0}\right]\) Velocity \([v]=\left[M^{0} L T^{-1}\right]\) Acceleration \([a]=\left[M^{0} L T^{-2}\right]\) and time period, \([T]=\left[M^{0} L^{0} T\right]\) Now, checking each option for these values For option(1) \(\dfrac{[a][T]}{[x]}=\dfrac{\left[M^{0} L T^{-2}\right]\left[M^{0} L^{0} T\right]}{\left[M^{0} L T^{0}\right]}=\left[M^{0} L^{0} T^{-1}\right]\) As it depends on time, so change with it. For option(2) \([a][T]+2 \pi[v]=\) \(\left[M^{0} L T^{-2}\right]\left[M^{0} L^{0} T\right]+\left[M^{0} L^{0} T^{-1}\right]\) \(=\left[M^{0} L T^{-1}\right]\) it is also dependent on time and hence changes with it For option (3) \(\begin{aligned}\dfrac{[a][T]}{[v]} & =\dfrac{\left[M^{0} L T^{-2}\right]\left[M^{0} L^{0} T\right]}{\left[M^{0} L T^{-1}\right]} \\& =\left[M^{0} L^{0} T^{0}\right]\end{aligned}\) As it is a constant having no dimension, so it does not change in time For option (4), \(\begin{aligned}& {[a][t]+4 \pi^{2}[d]^{2}=} \\& {\left[M^{0} L T^{-2}\right]\left[M^{0} L^{0} T\right]+\left[M^{0} L T^{-1}\right]^{2}} \\& =\left[L T^{-1}\right]+\left[L^{2} T^{-2}\right]\end{aligned}\) As, the term is dependent on time, so changes with it Also, it is dimensionally incorrect Hence option (3) is correct
364305
A body is executing simple harmonic motion with an angular frequency of \(2\,rad{s^{ - 1}}.\) The velocity of the body at \(20\;mm\) displacement, when the amplitude of the motion is \(60\;mm\) is
364306
The velocity of a particle in simple harmonic motion at displacement \(y\) from mean position is
1 \(\omega \sqrt{a^{2}-y^{2}}\)
2 \(\omega \sqrt{a^{2}+y^{2}}\)
3 \(\omega^{2} \sqrt{a^{2}-y^{2}}\)
4 \(\omega y\)
Explanation:
Conceptual Question
PHXI14:OSCILLATIONS
364307
Graph between velocity and displacement of a particle, executing S.H.M. is
1 A parabola
2 A straight line
3 An ellipse
4 A hyperbola
Explanation:
In simple harmonic motion \(y=a \sin \omega t\) and \(v=\omega \cos \omega t\) from this we have \(\dfrac{y^{2}}{a^{2}}+\dfrac{v^{2}}{a^{2} \omega^{2}}=1\), which is equation of ellipse.
PHXI14:OSCILLATIONS
364308
If ' \(x\) ', ' \(v\) 's, and a denote the displacement velocity and acceleration of a particle respectively executing SHM of preiodic time \(t\), then which one of the following does not change with time?
1 \(\dfrac{a T}{x}\)
2 \(a t+2 \pi v\)
3 \(\dfrac{a T}{v}\)
4 \(a t+4 \pi^{2} v^{2}\)
Explanation:
The dimensions of given variables of SHM are as Displacement \([x]=\left[M^{0} L T^{0}\right]\) Velocity \([v]=\left[M^{0} L T^{-1}\right]\) Acceleration \([a]=\left[M^{0} L T^{-2}\right]\) and time period, \([T]=\left[M^{0} L^{0} T\right]\) Now, checking each option for these values For option(1) \(\dfrac{[a][T]}{[x]}=\dfrac{\left[M^{0} L T^{-2}\right]\left[M^{0} L^{0} T\right]}{\left[M^{0} L T^{0}\right]}=\left[M^{0} L^{0} T^{-1}\right]\) As it depends on time, so change with it. For option(2) \([a][T]+2 \pi[v]=\) \(\left[M^{0} L T^{-2}\right]\left[M^{0} L^{0} T\right]+\left[M^{0} L^{0} T^{-1}\right]\) \(=\left[M^{0} L T^{-1}\right]\) it is also dependent on time and hence changes with it For option (3) \(\begin{aligned}\dfrac{[a][T]}{[v]} & =\dfrac{\left[M^{0} L T^{-2}\right]\left[M^{0} L^{0} T\right]}{\left[M^{0} L T^{-1}\right]} \\& =\left[M^{0} L^{0} T^{0}\right]\end{aligned}\) As it is a constant having no dimension, so it does not change in time For option (4), \(\begin{aligned}& {[a][t]+4 \pi^{2}[d]^{2}=} \\& {\left[M^{0} L T^{-2}\right]\left[M^{0} L^{0} T\right]+\left[M^{0} L T^{-1}\right]^{2}} \\& =\left[L T^{-1}\right]+\left[L^{2} T^{-2}\right]\end{aligned}\) As, the term is dependent on time, so changes with it Also, it is dimensionally incorrect Hence option (3) is correct
364305
A body is executing simple harmonic motion with an angular frequency of \(2\,rad{s^{ - 1}}.\) The velocity of the body at \(20\;mm\) displacement, when the amplitude of the motion is \(60\;mm\) is
364306
The velocity of a particle in simple harmonic motion at displacement \(y\) from mean position is
1 \(\omega \sqrt{a^{2}-y^{2}}\)
2 \(\omega \sqrt{a^{2}+y^{2}}\)
3 \(\omega^{2} \sqrt{a^{2}-y^{2}}\)
4 \(\omega y\)
Explanation:
Conceptual Question
PHXI14:OSCILLATIONS
364307
Graph between velocity and displacement of a particle, executing S.H.M. is
1 A parabola
2 A straight line
3 An ellipse
4 A hyperbola
Explanation:
In simple harmonic motion \(y=a \sin \omega t\) and \(v=\omega \cos \omega t\) from this we have \(\dfrac{y^{2}}{a^{2}}+\dfrac{v^{2}}{a^{2} \omega^{2}}=1\), which is equation of ellipse.
PHXI14:OSCILLATIONS
364308
If ' \(x\) ', ' \(v\) 's, and a denote the displacement velocity and acceleration of a particle respectively executing SHM of preiodic time \(t\), then which one of the following does not change with time?
1 \(\dfrac{a T}{x}\)
2 \(a t+2 \pi v\)
3 \(\dfrac{a T}{v}\)
4 \(a t+4 \pi^{2} v^{2}\)
Explanation:
The dimensions of given variables of SHM are as Displacement \([x]=\left[M^{0} L T^{0}\right]\) Velocity \([v]=\left[M^{0} L T^{-1}\right]\) Acceleration \([a]=\left[M^{0} L T^{-2}\right]\) and time period, \([T]=\left[M^{0} L^{0} T\right]\) Now, checking each option for these values For option(1) \(\dfrac{[a][T]}{[x]}=\dfrac{\left[M^{0} L T^{-2}\right]\left[M^{0} L^{0} T\right]}{\left[M^{0} L T^{0}\right]}=\left[M^{0} L^{0} T^{-1}\right]\) As it depends on time, so change with it. For option(2) \([a][T]+2 \pi[v]=\) \(\left[M^{0} L T^{-2}\right]\left[M^{0} L^{0} T\right]+\left[M^{0} L^{0} T^{-1}\right]\) \(=\left[M^{0} L T^{-1}\right]\) it is also dependent on time and hence changes with it For option (3) \(\begin{aligned}\dfrac{[a][T]}{[v]} & =\dfrac{\left[M^{0} L T^{-2}\right]\left[M^{0} L^{0} T\right]}{\left[M^{0} L T^{-1}\right]} \\& =\left[M^{0} L^{0} T^{0}\right]\end{aligned}\) As it is a constant having no dimension, so it does not change in time For option (4), \(\begin{aligned}& {[a][t]+4 \pi^{2}[d]^{2}=} \\& {\left[M^{0} L T^{-2}\right]\left[M^{0} L^{0} T\right]+\left[M^{0} L T^{-1}\right]^{2}} \\& =\left[L T^{-1}\right]+\left[L^{2} T^{-2}\right]\end{aligned}\) As, the term is dependent on time, so changes with it Also, it is dimensionally incorrect Hence option (3) is correct