140406
An elastic spring of unstretched length $L$ and force constant $K$ is stretched by a small length $x$. It is further stretched by another small length $y$. Work done during the second stretching is
D We know that, $\mathrm{W}=-\Delta \mathrm{U}$ Initial spring potential energy $\left(\mathrm{U}_{\mathrm{i}}\right)=\frac{1}{2} \mathrm{kx}^{2}$ Final spring potential energy $\left(\mathrm{U}_{\mathrm{f}}\right)=\frac{1}{2} \mathrm{k}(\mathrm{x}+\mathrm{y})^{2}$ Hence work done $(\mathrm{W})=\mathrm{U}_{\mathrm{i}}-\mathrm{U}_{\mathrm{f}}$ $(W)=\frac{1}{2} \mathrm{kx}^{2}-\frac{1}{2} \mathrm{k}(\mathrm{x}+\mathrm{y})^{2}$ $(\mathrm{~W})=\frac{1}{2} \mathrm{kx}^{2}-\frac{1}{2} \mathrm{k}\left(\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{xy}\right)$ $(\mathrm{W})=\frac{1}{2} \mathrm{kx}^{2}-\frac{1}{2} \mathrm{kx}^{2}-\frac{1}{2} \mathrm{ky}^{2}-\frac{1}{2} \mathrm{k}(2 \mathrm{xy})$ $(\mathrm{W})=-\frac{\mathrm{ky}}{2}(2 \mathrm{x}+\mathrm{y})$ Work done can not be negative Hence, $\mathrm{W}=\frac{\mathrm{ky}}{2}(2 \mathrm{x}+\mathrm{y})$
TS EAMCET (Engg.)-2016
Oscillations
140408
The potential energy of a simple harmonic oscillator when the particle is half way to its end point is
1 $\frac{1}{4} \mathrm{E}$
2 $\frac{1}{2} \mathrm{E}$
3 $\frac{2}{3} \mathrm{E}$
4 $\frac{1}{8} \mathrm{E}$
Explanation:
A Potential energy of a simple harmonic oscillator is, $=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{y}^{2}$ $\mathrm{PE}=\frac{1}{2} \mathrm{~m} \omega^{2}\left(\frac{\mathrm{a}}{2}\right)^{2}$ Kinetic energy of simple harmonic oscillator $\mathrm{KE}=\frac{1}{2} \mathrm{~m} \omega^{2}\left(\mathrm{~A}^{2}-\mathrm{y}^{2}\right)$ Total energy, $E=P E+K E$ $E=\frac{1}{2} m \omega^{2} y^{2}+\frac{1}{2} m \omega^{2}\left(A^{2}-y^{2}\right)$ $=\frac{1}{2} m \omega^{2} A^{2}$ $\text { For } y =\frac{\mathrm{a}}{2} \quad \text { (Given) }$ $=\frac{1}{2} \mathrm{~m} \omega^{2} \frac{\mathrm{a}^{2}}{4}$ $=\frac{1}{4}\left(\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{a}^{2}\right)=\frac{\mathrm{E}}{4}$
AIPMT-2003
Oscillations
140355
A vertical spring with force constant $k$ is fixed on a table. A ball of mass $m$ at a height $h$ above the free upper end of the spring falls vertically on the spring, so that the spring is compressed by a distance $d$. The net work done in the process is
B Work done is equal to change in energy of the body when mass $(\mathrm{m})$ falls vertically on spring, then spring is compressed by distance (d). $\therefore$ Net work done $(w)=$ Potential energy stored in the spring + Loss of potential energy $\mathrm{w}=\mathrm{mg}(\mathrm{h}+\mathrm{d})-\frac{1}{2} \mathrm{kd}^{2}$
AIIMS-2008
Oscillations
140362
A body executes simple harmonic motion with an amplitude $A$. At what displacement, from the mean position, is the potential energy of the body one fourth of its total energy?
1 $\frac{\mathrm{A}}{4}$
2 $\frac{\mathrm{A}}{2}$
3 $\frac{3 \mathrm{~A}}{4}$
4 $3 \mathrm{~A}$
Explanation:
B Given, potential energy (P.E.) $=\frac{1}{4}$ T.E. $\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{x}^{2} =\frac{1}{4}\left(\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}\right)$ $\mathrm{x}^{2} =\frac{\mathrm{A}^{2}}{4}$ $\mathrm{x} =\frac{\mathrm{A}}{2}$ Hence, the displacement from the mean position is $\mathrm{A} / 2$ when P.E. of the body one fourth of its total energy.
140406
An elastic spring of unstretched length $L$ and force constant $K$ is stretched by a small length $x$. It is further stretched by another small length $y$. Work done during the second stretching is
D We know that, $\mathrm{W}=-\Delta \mathrm{U}$ Initial spring potential energy $\left(\mathrm{U}_{\mathrm{i}}\right)=\frac{1}{2} \mathrm{kx}^{2}$ Final spring potential energy $\left(\mathrm{U}_{\mathrm{f}}\right)=\frac{1}{2} \mathrm{k}(\mathrm{x}+\mathrm{y})^{2}$ Hence work done $(\mathrm{W})=\mathrm{U}_{\mathrm{i}}-\mathrm{U}_{\mathrm{f}}$ $(W)=\frac{1}{2} \mathrm{kx}^{2}-\frac{1}{2} \mathrm{k}(\mathrm{x}+\mathrm{y})^{2}$ $(\mathrm{~W})=\frac{1}{2} \mathrm{kx}^{2}-\frac{1}{2} \mathrm{k}\left(\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{xy}\right)$ $(\mathrm{W})=\frac{1}{2} \mathrm{kx}^{2}-\frac{1}{2} \mathrm{kx}^{2}-\frac{1}{2} \mathrm{ky}^{2}-\frac{1}{2} \mathrm{k}(2 \mathrm{xy})$ $(\mathrm{W})=-\frac{\mathrm{ky}}{2}(2 \mathrm{x}+\mathrm{y})$ Work done can not be negative Hence, $\mathrm{W}=\frac{\mathrm{ky}}{2}(2 \mathrm{x}+\mathrm{y})$
TS EAMCET (Engg.)-2016
Oscillations
140408
The potential energy of a simple harmonic oscillator when the particle is half way to its end point is
1 $\frac{1}{4} \mathrm{E}$
2 $\frac{1}{2} \mathrm{E}$
3 $\frac{2}{3} \mathrm{E}$
4 $\frac{1}{8} \mathrm{E}$
Explanation:
A Potential energy of a simple harmonic oscillator is, $=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{y}^{2}$ $\mathrm{PE}=\frac{1}{2} \mathrm{~m} \omega^{2}\left(\frac{\mathrm{a}}{2}\right)^{2}$ Kinetic energy of simple harmonic oscillator $\mathrm{KE}=\frac{1}{2} \mathrm{~m} \omega^{2}\left(\mathrm{~A}^{2}-\mathrm{y}^{2}\right)$ Total energy, $E=P E+K E$ $E=\frac{1}{2} m \omega^{2} y^{2}+\frac{1}{2} m \omega^{2}\left(A^{2}-y^{2}\right)$ $=\frac{1}{2} m \omega^{2} A^{2}$ $\text { For } y =\frac{\mathrm{a}}{2} \quad \text { (Given) }$ $=\frac{1}{2} \mathrm{~m} \omega^{2} \frac{\mathrm{a}^{2}}{4}$ $=\frac{1}{4}\left(\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{a}^{2}\right)=\frac{\mathrm{E}}{4}$
AIPMT-2003
Oscillations
140355
A vertical spring with force constant $k$ is fixed on a table. A ball of mass $m$ at a height $h$ above the free upper end of the spring falls vertically on the spring, so that the spring is compressed by a distance $d$. The net work done in the process is
B Work done is equal to change in energy of the body when mass $(\mathrm{m})$ falls vertically on spring, then spring is compressed by distance (d). $\therefore$ Net work done $(w)=$ Potential energy stored in the spring + Loss of potential energy $\mathrm{w}=\mathrm{mg}(\mathrm{h}+\mathrm{d})-\frac{1}{2} \mathrm{kd}^{2}$
AIIMS-2008
Oscillations
140362
A body executes simple harmonic motion with an amplitude $A$. At what displacement, from the mean position, is the potential energy of the body one fourth of its total energy?
1 $\frac{\mathrm{A}}{4}$
2 $\frac{\mathrm{A}}{2}$
3 $\frac{3 \mathrm{~A}}{4}$
4 $3 \mathrm{~A}$
Explanation:
B Given, potential energy (P.E.) $=\frac{1}{4}$ T.E. $\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{x}^{2} =\frac{1}{4}\left(\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}\right)$ $\mathrm{x}^{2} =\frac{\mathrm{A}^{2}}{4}$ $\mathrm{x} =\frac{\mathrm{A}}{2}$ Hence, the displacement from the mean position is $\mathrm{A} / 2$ when P.E. of the body one fourth of its total energy.
140406
An elastic spring of unstretched length $L$ and force constant $K$ is stretched by a small length $x$. It is further stretched by another small length $y$. Work done during the second stretching is
D We know that, $\mathrm{W}=-\Delta \mathrm{U}$ Initial spring potential energy $\left(\mathrm{U}_{\mathrm{i}}\right)=\frac{1}{2} \mathrm{kx}^{2}$ Final spring potential energy $\left(\mathrm{U}_{\mathrm{f}}\right)=\frac{1}{2} \mathrm{k}(\mathrm{x}+\mathrm{y})^{2}$ Hence work done $(\mathrm{W})=\mathrm{U}_{\mathrm{i}}-\mathrm{U}_{\mathrm{f}}$ $(W)=\frac{1}{2} \mathrm{kx}^{2}-\frac{1}{2} \mathrm{k}(\mathrm{x}+\mathrm{y})^{2}$ $(\mathrm{~W})=\frac{1}{2} \mathrm{kx}^{2}-\frac{1}{2} \mathrm{k}\left(\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{xy}\right)$ $(\mathrm{W})=\frac{1}{2} \mathrm{kx}^{2}-\frac{1}{2} \mathrm{kx}^{2}-\frac{1}{2} \mathrm{ky}^{2}-\frac{1}{2} \mathrm{k}(2 \mathrm{xy})$ $(\mathrm{W})=-\frac{\mathrm{ky}}{2}(2 \mathrm{x}+\mathrm{y})$ Work done can not be negative Hence, $\mathrm{W}=\frac{\mathrm{ky}}{2}(2 \mathrm{x}+\mathrm{y})$
TS EAMCET (Engg.)-2016
Oscillations
140408
The potential energy of a simple harmonic oscillator when the particle is half way to its end point is
1 $\frac{1}{4} \mathrm{E}$
2 $\frac{1}{2} \mathrm{E}$
3 $\frac{2}{3} \mathrm{E}$
4 $\frac{1}{8} \mathrm{E}$
Explanation:
A Potential energy of a simple harmonic oscillator is, $=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{y}^{2}$ $\mathrm{PE}=\frac{1}{2} \mathrm{~m} \omega^{2}\left(\frac{\mathrm{a}}{2}\right)^{2}$ Kinetic energy of simple harmonic oscillator $\mathrm{KE}=\frac{1}{2} \mathrm{~m} \omega^{2}\left(\mathrm{~A}^{2}-\mathrm{y}^{2}\right)$ Total energy, $E=P E+K E$ $E=\frac{1}{2} m \omega^{2} y^{2}+\frac{1}{2} m \omega^{2}\left(A^{2}-y^{2}\right)$ $=\frac{1}{2} m \omega^{2} A^{2}$ $\text { For } y =\frac{\mathrm{a}}{2} \quad \text { (Given) }$ $=\frac{1}{2} \mathrm{~m} \omega^{2} \frac{\mathrm{a}^{2}}{4}$ $=\frac{1}{4}\left(\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{a}^{2}\right)=\frac{\mathrm{E}}{4}$
AIPMT-2003
Oscillations
140355
A vertical spring with force constant $k$ is fixed on a table. A ball of mass $m$ at a height $h$ above the free upper end of the spring falls vertically on the spring, so that the spring is compressed by a distance $d$. The net work done in the process is
B Work done is equal to change in energy of the body when mass $(\mathrm{m})$ falls vertically on spring, then spring is compressed by distance (d). $\therefore$ Net work done $(w)=$ Potential energy stored in the spring + Loss of potential energy $\mathrm{w}=\mathrm{mg}(\mathrm{h}+\mathrm{d})-\frac{1}{2} \mathrm{kd}^{2}$
AIIMS-2008
Oscillations
140362
A body executes simple harmonic motion with an amplitude $A$. At what displacement, from the mean position, is the potential energy of the body one fourth of its total energy?
1 $\frac{\mathrm{A}}{4}$
2 $\frac{\mathrm{A}}{2}$
3 $\frac{3 \mathrm{~A}}{4}$
4 $3 \mathrm{~A}$
Explanation:
B Given, potential energy (P.E.) $=\frac{1}{4}$ T.E. $\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{x}^{2} =\frac{1}{4}\left(\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}\right)$ $\mathrm{x}^{2} =\frac{\mathrm{A}^{2}}{4}$ $\mathrm{x} =\frac{\mathrm{A}}{2}$ Hence, the displacement from the mean position is $\mathrm{A} / 2$ when P.E. of the body one fourth of its total energy.
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Oscillations
140406
An elastic spring of unstretched length $L$ and force constant $K$ is stretched by a small length $x$. It is further stretched by another small length $y$. Work done during the second stretching is
D We know that, $\mathrm{W}=-\Delta \mathrm{U}$ Initial spring potential energy $\left(\mathrm{U}_{\mathrm{i}}\right)=\frac{1}{2} \mathrm{kx}^{2}$ Final spring potential energy $\left(\mathrm{U}_{\mathrm{f}}\right)=\frac{1}{2} \mathrm{k}(\mathrm{x}+\mathrm{y})^{2}$ Hence work done $(\mathrm{W})=\mathrm{U}_{\mathrm{i}}-\mathrm{U}_{\mathrm{f}}$ $(W)=\frac{1}{2} \mathrm{kx}^{2}-\frac{1}{2} \mathrm{k}(\mathrm{x}+\mathrm{y})^{2}$ $(\mathrm{~W})=\frac{1}{2} \mathrm{kx}^{2}-\frac{1}{2} \mathrm{k}\left(\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{xy}\right)$ $(\mathrm{W})=\frac{1}{2} \mathrm{kx}^{2}-\frac{1}{2} \mathrm{kx}^{2}-\frac{1}{2} \mathrm{ky}^{2}-\frac{1}{2} \mathrm{k}(2 \mathrm{xy})$ $(\mathrm{W})=-\frac{\mathrm{ky}}{2}(2 \mathrm{x}+\mathrm{y})$ Work done can not be negative Hence, $\mathrm{W}=\frac{\mathrm{ky}}{2}(2 \mathrm{x}+\mathrm{y})$
TS EAMCET (Engg.)-2016
Oscillations
140408
The potential energy of a simple harmonic oscillator when the particle is half way to its end point is
1 $\frac{1}{4} \mathrm{E}$
2 $\frac{1}{2} \mathrm{E}$
3 $\frac{2}{3} \mathrm{E}$
4 $\frac{1}{8} \mathrm{E}$
Explanation:
A Potential energy of a simple harmonic oscillator is, $=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{y}^{2}$ $\mathrm{PE}=\frac{1}{2} \mathrm{~m} \omega^{2}\left(\frac{\mathrm{a}}{2}\right)^{2}$ Kinetic energy of simple harmonic oscillator $\mathrm{KE}=\frac{1}{2} \mathrm{~m} \omega^{2}\left(\mathrm{~A}^{2}-\mathrm{y}^{2}\right)$ Total energy, $E=P E+K E$ $E=\frac{1}{2} m \omega^{2} y^{2}+\frac{1}{2} m \omega^{2}\left(A^{2}-y^{2}\right)$ $=\frac{1}{2} m \omega^{2} A^{2}$ $\text { For } y =\frac{\mathrm{a}}{2} \quad \text { (Given) }$ $=\frac{1}{2} \mathrm{~m} \omega^{2} \frac{\mathrm{a}^{2}}{4}$ $=\frac{1}{4}\left(\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{a}^{2}\right)=\frac{\mathrm{E}}{4}$
AIPMT-2003
Oscillations
140355
A vertical spring with force constant $k$ is fixed on a table. A ball of mass $m$ at a height $h$ above the free upper end of the spring falls vertically on the spring, so that the spring is compressed by a distance $d$. The net work done in the process is
B Work done is equal to change in energy of the body when mass $(\mathrm{m})$ falls vertically on spring, then spring is compressed by distance (d). $\therefore$ Net work done $(w)=$ Potential energy stored in the spring + Loss of potential energy $\mathrm{w}=\mathrm{mg}(\mathrm{h}+\mathrm{d})-\frac{1}{2} \mathrm{kd}^{2}$
AIIMS-2008
Oscillations
140362
A body executes simple harmonic motion with an amplitude $A$. At what displacement, from the mean position, is the potential energy of the body one fourth of its total energy?
1 $\frac{\mathrm{A}}{4}$
2 $\frac{\mathrm{A}}{2}$
3 $\frac{3 \mathrm{~A}}{4}$
4 $3 \mathrm{~A}$
Explanation:
B Given, potential energy (P.E.) $=\frac{1}{4}$ T.E. $\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{x}^{2} =\frac{1}{4}\left(\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}\right)$ $\mathrm{x}^{2} =\frac{\mathrm{A}^{2}}{4}$ $\mathrm{x} =\frac{\mathrm{A}}{2}$ Hence, the displacement from the mean position is $\mathrm{A} / 2$ when P.E. of the body one fourth of its total energy.
