140376
A spring with force constant $k$ is initially stretched by $x_{1}$. If it is further stretched by $x_{2}$, then the increase in its potential energy is
B According to the question, First condition potential energy $\mathrm{U}_{1}=\frac{1}{2} \mathrm{kx}_{1}^{2}$ In second condition $\mathrm{U}_{2}=\frac{1}{2} \mathrm{k}\left(\mathrm{x}_{1}+\mathrm{x}_{2}\right)^{2}$ Increase in the potential energy $\Delta \mathrm{U} =\mathrm{U}_{2}-\mathrm{U}_{1}$ $=\frac{1}{2} \mathrm{k}\left(\mathrm{x}_{1}+\mathrm{x}_{2}\right)^{2}-\frac{1}{2} \mathrm{kx}_{1}^{2}$ $=\frac{1}{2} \mathrm{k}\left[\mathrm{x}_{1}^{2}+\mathrm{x}_{2}^{2}+2 \mathrm{x}_{1} \mathrm{x}_{2}-\mathrm{x}_{1}^{2}\right]$ $=\frac{1}{2} \mathrm{k}\left[2 \mathrm{x}_{1} \mathrm{x}_{2}+\mathrm{x}_{2}^{2}\right]$ $\Delta \mathrm{U} =\frac{1}{2} \mathrm{kx}_{2}\left[\mathrm{x}_{2}+2 \mathrm{x}_{1}\right]$
Kerala CEE - 2015
Oscillations
140378
Two identical pendulums are oscillating with amplitudes $4 \mathrm{~cm}$ and $8 \mathrm{~cm}$. The ratio of their energies of oscillation will be
1 $1 / 3$
2 $1 / 4$
3 $1 / 9$
4 $1 / 2$
Explanation:
B Given that, $\mathrm{m}_{1}=\mathrm{m}_{2}=\mathrm{m}, \omega_{1}=\omega_{2}=\omega, \mathrm{A}_{1}=$ $4 \mathrm{~cm}, \mathrm{~A}_{2}=8 \mathrm{~cm}$ Total energies of oscillation $\mathrm{E}=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}$ Then, $\quad \mathrm{E}_{1}=\frac{1}{2} \mathrm{~m} \omega^{2}(4)^{2}$ $\mathrm{E}_{2}=\frac{1}{2} \mathrm{~m} \omega^{2}(8)^{2}$ $\therefore \quad \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{4}{8}\right)^{2}=\frac{1}{4}$
Oscillations
140381
A body executes SHM. At a displacement $x$, its potential energy is $E_{1}$. At a displacement $y$, its potential energy is $\mathbf{E}_{2}$. What is the potential energy of the body at a displacement $(x+y)$ ?
140376
A spring with force constant $k$ is initially stretched by $x_{1}$. If it is further stretched by $x_{2}$, then the increase in its potential energy is
B According to the question, First condition potential energy $\mathrm{U}_{1}=\frac{1}{2} \mathrm{kx}_{1}^{2}$ In second condition $\mathrm{U}_{2}=\frac{1}{2} \mathrm{k}\left(\mathrm{x}_{1}+\mathrm{x}_{2}\right)^{2}$ Increase in the potential energy $\Delta \mathrm{U} =\mathrm{U}_{2}-\mathrm{U}_{1}$ $=\frac{1}{2} \mathrm{k}\left(\mathrm{x}_{1}+\mathrm{x}_{2}\right)^{2}-\frac{1}{2} \mathrm{kx}_{1}^{2}$ $=\frac{1}{2} \mathrm{k}\left[\mathrm{x}_{1}^{2}+\mathrm{x}_{2}^{2}+2 \mathrm{x}_{1} \mathrm{x}_{2}-\mathrm{x}_{1}^{2}\right]$ $=\frac{1}{2} \mathrm{k}\left[2 \mathrm{x}_{1} \mathrm{x}_{2}+\mathrm{x}_{2}^{2}\right]$ $\Delta \mathrm{U} =\frac{1}{2} \mathrm{kx}_{2}\left[\mathrm{x}_{2}+2 \mathrm{x}_{1}\right]$
Kerala CEE - 2015
Oscillations
140378
Two identical pendulums are oscillating with amplitudes $4 \mathrm{~cm}$ and $8 \mathrm{~cm}$. The ratio of their energies of oscillation will be
1 $1 / 3$
2 $1 / 4$
3 $1 / 9$
4 $1 / 2$
Explanation:
B Given that, $\mathrm{m}_{1}=\mathrm{m}_{2}=\mathrm{m}, \omega_{1}=\omega_{2}=\omega, \mathrm{A}_{1}=$ $4 \mathrm{~cm}, \mathrm{~A}_{2}=8 \mathrm{~cm}$ Total energies of oscillation $\mathrm{E}=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}$ Then, $\quad \mathrm{E}_{1}=\frac{1}{2} \mathrm{~m} \omega^{2}(4)^{2}$ $\mathrm{E}_{2}=\frac{1}{2} \mathrm{~m} \omega^{2}(8)^{2}$ $\therefore \quad \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{4}{8}\right)^{2}=\frac{1}{4}$
Oscillations
140381
A body executes SHM. At a displacement $x$, its potential energy is $E_{1}$. At a displacement $y$, its potential energy is $\mathbf{E}_{2}$. What is the potential energy of the body at a displacement $(x+y)$ ?
140376
A spring with force constant $k$ is initially stretched by $x_{1}$. If it is further stretched by $x_{2}$, then the increase in its potential energy is
B According to the question, First condition potential energy $\mathrm{U}_{1}=\frac{1}{2} \mathrm{kx}_{1}^{2}$ In second condition $\mathrm{U}_{2}=\frac{1}{2} \mathrm{k}\left(\mathrm{x}_{1}+\mathrm{x}_{2}\right)^{2}$ Increase in the potential energy $\Delta \mathrm{U} =\mathrm{U}_{2}-\mathrm{U}_{1}$ $=\frac{1}{2} \mathrm{k}\left(\mathrm{x}_{1}+\mathrm{x}_{2}\right)^{2}-\frac{1}{2} \mathrm{kx}_{1}^{2}$ $=\frac{1}{2} \mathrm{k}\left[\mathrm{x}_{1}^{2}+\mathrm{x}_{2}^{2}+2 \mathrm{x}_{1} \mathrm{x}_{2}-\mathrm{x}_{1}^{2}\right]$ $=\frac{1}{2} \mathrm{k}\left[2 \mathrm{x}_{1} \mathrm{x}_{2}+\mathrm{x}_{2}^{2}\right]$ $\Delta \mathrm{U} =\frac{1}{2} \mathrm{kx}_{2}\left[\mathrm{x}_{2}+2 \mathrm{x}_{1}\right]$
Kerala CEE - 2015
Oscillations
140378
Two identical pendulums are oscillating with amplitudes $4 \mathrm{~cm}$ and $8 \mathrm{~cm}$. The ratio of their energies of oscillation will be
1 $1 / 3$
2 $1 / 4$
3 $1 / 9$
4 $1 / 2$
Explanation:
B Given that, $\mathrm{m}_{1}=\mathrm{m}_{2}=\mathrm{m}, \omega_{1}=\omega_{2}=\omega, \mathrm{A}_{1}=$ $4 \mathrm{~cm}, \mathrm{~A}_{2}=8 \mathrm{~cm}$ Total energies of oscillation $\mathrm{E}=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}$ Then, $\quad \mathrm{E}_{1}=\frac{1}{2} \mathrm{~m} \omega^{2}(4)^{2}$ $\mathrm{E}_{2}=\frac{1}{2} \mathrm{~m} \omega^{2}(8)^{2}$ $\therefore \quad \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{4}{8}\right)^{2}=\frac{1}{4}$
Oscillations
140381
A body executes SHM. At a displacement $x$, its potential energy is $E_{1}$. At a displacement $y$, its potential energy is $\mathbf{E}_{2}$. What is the potential energy of the body at a displacement $(x+y)$ ?
140376
A spring with force constant $k$ is initially stretched by $x_{1}$. If it is further stretched by $x_{2}$, then the increase in its potential energy is
B According to the question, First condition potential energy $\mathrm{U}_{1}=\frac{1}{2} \mathrm{kx}_{1}^{2}$ In second condition $\mathrm{U}_{2}=\frac{1}{2} \mathrm{k}\left(\mathrm{x}_{1}+\mathrm{x}_{2}\right)^{2}$ Increase in the potential energy $\Delta \mathrm{U} =\mathrm{U}_{2}-\mathrm{U}_{1}$ $=\frac{1}{2} \mathrm{k}\left(\mathrm{x}_{1}+\mathrm{x}_{2}\right)^{2}-\frac{1}{2} \mathrm{kx}_{1}^{2}$ $=\frac{1}{2} \mathrm{k}\left[\mathrm{x}_{1}^{2}+\mathrm{x}_{2}^{2}+2 \mathrm{x}_{1} \mathrm{x}_{2}-\mathrm{x}_{1}^{2}\right]$ $=\frac{1}{2} \mathrm{k}\left[2 \mathrm{x}_{1} \mathrm{x}_{2}+\mathrm{x}_{2}^{2}\right]$ $\Delta \mathrm{U} =\frac{1}{2} \mathrm{kx}_{2}\left[\mathrm{x}_{2}+2 \mathrm{x}_{1}\right]$
Kerala CEE - 2015
Oscillations
140378
Two identical pendulums are oscillating with amplitudes $4 \mathrm{~cm}$ and $8 \mathrm{~cm}$. The ratio of their energies of oscillation will be
1 $1 / 3$
2 $1 / 4$
3 $1 / 9$
4 $1 / 2$
Explanation:
B Given that, $\mathrm{m}_{1}=\mathrm{m}_{2}=\mathrm{m}, \omega_{1}=\omega_{2}=\omega, \mathrm{A}_{1}=$ $4 \mathrm{~cm}, \mathrm{~A}_{2}=8 \mathrm{~cm}$ Total energies of oscillation $\mathrm{E}=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}$ Then, $\quad \mathrm{E}_{1}=\frac{1}{2} \mathrm{~m} \omega^{2}(4)^{2}$ $\mathrm{E}_{2}=\frac{1}{2} \mathrm{~m} \omega^{2}(8)^{2}$ $\therefore \quad \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{4}{8}\right)^{2}=\frac{1}{4}$
Oscillations
140381
A body executes SHM. At a displacement $x$, its potential energy is $E_{1}$. At a displacement $y$, its potential energy is $\mathbf{E}_{2}$. What is the potential energy of the body at a displacement $(x+y)$ ?
