364141
The potential energy of a particle \(\left( {{U_x}} \right)\) executing \(S.H.M.\) is given by
1 \({U_x} = \frac{k}{2}{(x - a)^2}\)
2 \({U_x} = {k_1}x + {k_2}{x^2} + {k_3}{x^3}\)
3 \({U_x} = A{e^{ - bx}}\)
4 \({U_x} = {\rm{a }}\,{\rm{constant}}\)
Explanation:
\(P.E.\) of body in \(S.H.M.\) at an instant, \(U=\dfrac{1}{2} m \omega^{2} y^{2}=\dfrac{1}{2} k y^{2}\) If the displacement, \(y=(a-x)\) then$ U=\dfrac{1}{2} k(a-x)^{2}=\dfrac{1}{2} k(x-a)^{2}$
PHXI14:OSCILLATIONS
364142
The average kinetic energy of a simple harmonic oscillator is 2 joule and its total energy is 5 joule. Its minimum potential energy is
364143
The total energy of a simple harmonic oscillator is proportional to
1 Square of the amplitude
2 Square root of displacement
3 Amplitude
4 Frequency
Explanation:
Total energy of simple harmonic oscillator is given by \(E=\dfrac{1}{2} m \omega^{2} A^{2}\) where, \(m = \) mass of body performing SHM \(\omega=\) angular velocity and \(A = \) amplitude \(\therefore E \propto A^{2}\)
MHTCET - 2019
PHXI14:OSCILLATIONS
364144
The physical quantity conserved in simple harmonic motion is
1 time period
2 total energy
3 displacement
4 force
Explanation:
Total energy remains conserved in simple harmonic motion. So correct option is (2)
PHXI14:OSCILLATIONS
364145
The displacement of a particle of mass \(3\,gm\) executing simple harmonic motion is given by \(y=3 \sin (0.2 t)\) in SI units. The kinetic energy of the particle at a point which is at a distance equal to \(\dfrac{1}{3}\) of its amplitude from its mean position is
364141
The potential energy of a particle \(\left( {{U_x}} \right)\) executing \(S.H.M.\) is given by
1 \({U_x} = \frac{k}{2}{(x - a)^2}\)
2 \({U_x} = {k_1}x + {k_2}{x^2} + {k_3}{x^3}\)
3 \({U_x} = A{e^{ - bx}}\)
4 \({U_x} = {\rm{a }}\,{\rm{constant}}\)
Explanation:
\(P.E.\) of body in \(S.H.M.\) at an instant, \(U=\dfrac{1}{2} m \omega^{2} y^{2}=\dfrac{1}{2} k y^{2}\) If the displacement, \(y=(a-x)\) then$ U=\dfrac{1}{2} k(a-x)^{2}=\dfrac{1}{2} k(x-a)^{2}$
PHXI14:OSCILLATIONS
364142
The average kinetic energy of a simple harmonic oscillator is 2 joule and its total energy is 5 joule. Its minimum potential energy is
364143
The total energy of a simple harmonic oscillator is proportional to
1 Square of the amplitude
2 Square root of displacement
3 Amplitude
4 Frequency
Explanation:
Total energy of simple harmonic oscillator is given by \(E=\dfrac{1}{2} m \omega^{2} A^{2}\) where, \(m = \) mass of body performing SHM \(\omega=\) angular velocity and \(A = \) amplitude \(\therefore E \propto A^{2}\)
MHTCET - 2019
PHXI14:OSCILLATIONS
364144
The physical quantity conserved in simple harmonic motion is
1 time period
2 total energy
3 displacement
4 force
Explanation:
Total energy remains conserved in simple harmonic motion. So correct option is (2)
PHXI14:OSCILLATIONS
364145
The displacement of a particle of mass \(3\,gm\) executing simple harmonic motion is given by \(y=3 \sin (0.2 t)\) in SI units. The kinetic energy of the particle at a point which is at a distance equal to \(\dfrac{1}{3}\) of its amplitude from its mean position is
364141
The potential energy of a particle \(\left( {{U_x}} \right)\) executing \(S.H.M.\) is given by
1 \({U_x} = \frac{k}{2}{(x - a)^2}\)
2 \({U_x} = {k_1}x + {k_2}{x^2} + {k_3}{x^3}\)
3 \({U_x} = A{e^{ - bx}}\)
4 \({U_x} = {\rm{a }}\,{\rm{constant}}\)
Explanation:
\(P.E.\) of body in \(S.H.M.\) at an instant, \(U=\dfrac{1}{2} m \omega^{2} y^{2}=\dfrac{1}{2} k y^{2}\) If the displacement, \(y=(a-x)\) then$ U=\dfrac{1}{2} k(a-x)^{2}=\dfrac{1}{2} k(x-a)^{2}$
PHXI14:OSCILLATIONS
364142
The average kinetic energy of a simple harmonic oscillator is 2 joule and its total energy is 5 joule. Its minimum potential energy is
364143
The total energy of a simple harmonic oscillator is proportional to
1 Square of the amplitude
2 Square root of displacement
3 Amplitude
4 Frequency
Explanation:
Total energy of simple harmonic oscillator is given by \(E=\dfrac{1}{2} m \omega^{2} A^{2}\) where, \(m = \) mass of body performing SHM \(\omega=\) angular velocity and \(A = \) amplitude \(\therefore E \propto A^{2}\)
MHTCET - 2019
PHXI14:OSCILLATIONS
364144
The physical quantity conserved in simple harmonic motion is
1 time period
2 total energy
3 displacement
4 force
Explanation:
Total energy remains conserved in simple harmonic motion. So correct option is (2)
PHXI14:OSCILLATIONS
364145
The displacement of a particle of mass \(3\,gm\) executing simple harmonic motion is given by \(y=3 \sin (0.2 t)\) in SI units. The kinetic energy of the particle at a point which is at a distance equal to \(\dfrac{1}{3}\) of its amplitude from its mean position is
364141
The potential energy of a particle \(\left( {{U_x}} \right)\) executing \(S.H.M.\) is given by
1 \({U_x} = \frac{k}{2}{(x - a)^2}\)
2 \({U_x} = {k_1}x + {k_2}{x^2} + {k_3}{x^3}\)
3 \({U_x} = A{e^{ - bx}}\)
4 \({U_x} = {\rm{a }}\,{\rm{constant}}\)
Explanation:
\(P.E.\) of body in \(S.H.M.\) at an instant, \(U=\dfrac{1}{2} m \omega^{2} y^{2}=\dfrac{1}{2} k y^{2}\) If the displacement, \(y=(a-x)\) then$ U=\dfrac{1}{2} k(a-x)^{2}=\dfrac{1}{2} k(x-a)^{2}$
PHXI14:OSCILLATIONS
364142
The average kinetic energy of a simple harmonic oscillator is 2 joule and its total energy is 5 joule. Its minimum potential energy is
364143
The total energy of a simple harmonic oscillator is proportional to
1 Square of the amplitude
2 Square root of displacement
3 Amplitude
4 Frequency
Explanation:
Total energy of simple harmonic oscillator is given by \(E=\dfrac{1}{2} m \omega^{2} A^{2}\) where, \(m = \) mass of body performing SHM \(\omega=\) angular velocity and \(A = \) amplitude \(\therefore E \propto A^{2}\)
MHTCET - 2019
PHXI14:OSCILLATIONS
364144
The physical quantity conserved in simple harmonic motion is
1 time period
2 total energy
3 displacement
4 force
Explanation:
Total energy remains conserved in simple harmonic motion. So correct option is (2)
PHXI14:OSCILLATIONS
364145
The displacement of a particle of mass \(3\,gm\) executing simple harmonic motion is given by \(y=3 \sin (0.2 t)\) in SI units. The kinetic energy of the particle at a point which is at a distance equal to \(\dfrac{1}{3}\) of its amplitude from its mean position is
364141
The potential energy of a particle \(\left( {{U_x}} \right)\) executing \(S.H.M.\) is given by
1 \({U_x} = \frac{k}{2}{(x - a)^2}\)
2 \({U_x} = {k_1}x + {k_2}{x^2} + {k_3}{x^3}\)
3 \({U_x} = A{e^{ - bx}}\)
4 \({U_x} = {\rm{a }}\,{\rm{constant}}\)
Explanation:
\(P.E.\) of body in \(S.H.M.\) at an instant, \(U=\dfrac{1}{2} m \omega^{2} y^{2}=\dfrac{1}{2} k y^{2}\) If the displacement, \(y=(a-x)\) then$ U=\dfrac{1}{2} k(a-x)^{2}=\dfrac{1}{2} k(x-a)^{2}$
PHXI14:OSCILLATIONS
364142
The average kinetic energy of a simple harmonic oscillator is 2 joule and its total energy is 5 joule. Its minimum potential energy is
364143
The total energy of a simple harmonic oscillator is proportional to
1 Square of the amplitude
2 Square root of displacement
3 Amplitude
4 Frequency
Explanation:
Total energy of simple harmonic oscillator is given by \(E=\dfrac{1}{2} m \omega^{2} A^{2}\) where, \(m = \) mass of body performing SHM \(\omega=\) angular velocity and \(A = \) amplitude \(\therefore E \propto A^{2}\)
MHTCET - 2019
PHXI14:OSCILLATIONS
364144
The physical quantity conserved in simple harmonic motion is
1 time period
2 total energy
3 displacement
4 force
Explanation:
Total energy remains conserved in simple harmonic motion. So correct option is (2)
PHXI14:OSCILLATIONS
364145
The displacement of a particle of mass \(3\,gm\) executing simple harmonic motion is given by \(y=3 \sin (0.2 t)\) in SI units. The kinetic energy of the particle at a point which is at a distance equal to \(\dfrac{1}{3}\) of its amplitude from its mean position is
