140367
The angular retardation of a rotating flywheel is proportional to the angle through which it rotates. If its kinetic energy gets reduced by $\Delta \mathrm{E}$ while it rotates through at an angle $\theta$, then
A The condition of S.H.M is $-\mathrm{a} \propto \theta$ We know that, $\mathrm{f}_{\mathrm{r}}=-$ ma $\mathrm{f}_{\mathrm{r}} \propto \theta$ $\mathrm{f}_{\mathrm{r}}=\mathrm{k} \theta$ Law of conservation of energy- $\Delta \mathrm{W} =\Delta \mathrm{KE}$ $\Delta \mathrm{E} =\mathrm{f}_{\mathrm{r}} \cdot \theta$ $\Delta \mathrm{E} =\mathrm{k} \cdot \theta \cdot \theta \quad\left(\mathrm{f}_{\mathrm{r}}=\mathrm{k} \theta\right)$ $\Delta \mathrm{E} =\mathrm{k} \theta^{2}$ Hence, reduced kinetic energy $(\Delta \mathrm{E}) \propto \theta^{2}$
AP EAMCET (21.04.2019) Shift-II
Oscillations
140368
A particle starts oscillating in simple harmonic motion form its equilibrium position with time period $T$. The ratio of $K E$ and $P E$ of the particle at time $\mathrm{t}=\frac{\mathrm{T}}{12}$ is
1 $3: 1$
2 $1: 4$
3 $4: 1$
4 $2: 1$
Explanation:
A We know that, displacement equation in S.H.M is given by- $\mathrm{x}=\mathrm{a} \sin \omega \mathrm{t}, \quad \mathrm{t}=\frac{\mathrm{T}}{12}$ $x=a \sin \left(\omega \times \frac{T}{12}\right)$ $x=a \sin \left(\frac{2 \pi}{T} \times \frac{T}{12}\right)$ $x=a \sin 30^{\circ}$ $x=\frac{a}{2}$ Kinetic energy of the particle, $K E=\frac{1}{2} k\left(a^{2}-x^{2}\right)$ Potential energy $(P E)=\frac{1}{2} \mathrm{kx}^{2}$ $\therefore \quad \quad \frac{K . E}{P . E} =\frac{\frac{1}{2} \mathrm{k}\left(\mathrm{a}^{2}-\mathrm{x}^{2}\right)}{\frac{1}{2} \mathrm{kx}^{2}}=\frac{\mathrm{a}^{2}-\mathrm{x}^{2}}{\mathrm{x}^{2}}$ $\frac{\mathrm{K} . \mathrm{E}}{\mathrm{P} . \mathrm{E}} =\frac{\mathrm{a}^{2}-\left(\frac{\mathrm{a}}{2}\right)^{2}}{\left(\frac{\mathrm{a}}{2}\right)^{2}}=\frac{\frac{4 \mathrm{a}^{2}-\mathrm{a}^{2}}{4}}{\frac{\mathrm{a}^{2}}{4}}=\frac{\frac{3 \mathrm{a}^{2}}{4}}{\frac{\mathrm{a}^{2}}{4}}=\frac{3 \mathrm{a}^{2}}{\mathrm{a}^{2}}$ $\frac{\mathrm{K} . \mathrm{E}}{\mathrm{P} . \mathrm{E}} =\frac{3}{1}$
CG PET 2019
Oscillations
140369
Assertion: If the amplitude of a simple harmonic oscillator is doubled, its total energy becomes four times. Reason: The total energy is directly proportional to the square of the amplitude of vibration of the harmonic oscillator.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason in incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
A Total mechanical energy of a simple harmonic oscillator $\mathrm{E}=\frac{1}{2} \mathrm{ma}^{2} \omega^{2}$ $\mathrm{E} \propto \mathrm{a}^{2}$ $\therefore \quad \frac{E_{1}}{E_{2}}=\left(\frac{a_{1}}{a_{2}}\right)^{2}$ Given that, $a_{1}=a, a_{2}=2 a$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{\mathrm{a}}{2 \mathrm{a}}\right)^{2}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{1}{4}$ $4 \mathrm{E}_{1}=\mathrm{E}_{2}$
AIIMS-26.05.2018(E)
Oscillations
140370
The ratio of displacement to amplitude, when kinetic energy of a body executing SHM is thrice the potential energy
1 $\frac{3}{2}$
2 $\frac{4}{3}$
3 $\frac{1}{2}$
4 $\frac{2}{3}$
Explanation:
C Given that, K.E. $=3 \mathrm{U}$ We know that, K.E. in SHM- $\text { K.E. }=\frac{1}{2} m \omega^{2}\left(a^{2}-x^{2}\right)$ Potential energy in S.H.M. $\mathrm{U}=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{x}^{2}$ From equation (i), we get- $\frac{1}{2} m \omega^{2}\left(a^{2}-x^{2}\right)=3 \times \frac{1}{2} m \omega^{2} x^{2}$ $a^{2}-x^{2}=3 x^{2}$ $4 x^{2}=a^{2}$ $a=2 x$ $\frac{x}{a}=\frac{1}{2}$
140367
The angular retardation of a rotating flywheel is proportional to the angle through which it rotates. If its kinetic energy gets reduced by $\Delta \mathrm{E}$ while it rotates through at an angle $\theta$, then
A The condition of S.H.M is $-\mathrm{a} \propto \theta$ We know that, $\mathrm{f}_{\mathrm{r}}=-$ ma $\mathrm{f}_{\mathrm{r}} \propto \theta$ $\mathrm{f}_{\mathrm{r}}=\mathrm{k} \theta$ Law of conservation of energy- $\Delta \mathrm{W} =\Delta \mathrm{KE}$ $\Delta \mathrm{E} =\mathrm{f}_{\mathrm{r}} \cdot \theta$ $\Delta \mathrm{E} =\mathrm{k} \cdot \theta \cdot \theta \quad\left(\mathrm{f}_{\mathrm{r}}=\mathrm{k} \theta\right)$ $\Delta \mathrm{E} =\mathrm{k} \theta^{2}$ Hence, reduced kinetic energy $(\Delta \mathrm{E}) \propto \theta^{2}$
AP EAMCET (21.04.2019) Shift-II
Oscillations
140368
A particle starts oscillating in simple harmonic motion form its equilibrium position with time period $T$. The ratio of $K E$ and $P E$ of the particle at time $\mathrm{t}=\frac{\mathrm{T}}{12}$ is
1 $3: 1$
2 $1: 4$
3 $4: 1$
4 $2: 1$
Explanation:
A We know that, displacement equation in S.H.M is given by- $\mathrm{x}=\mathrm{a} \sin \omega \mathrm{t}, \quad \mathrm{t}=\frac{\mathrm{T}}{12}$ $x=a \sin \left(\omega \times \frac{T}{12}\right)$ $x=a \sin \left(\frac{2 \pi}{T} \times \frac{T}{12}\right)$ $x=a \sin 30^{\circ}$ $x=\frac{a}{2}$ Kinetic energy of the particle, $K E=\frac{1}{2} k\left(a^{2}-x^{2}\right)$ Potential energy $(P E)=\frac{1}{2} \mathrm{kx}^{2}$ $\therefore \quad \quad \frac{K . E}{P . E} =\frac{\frac{1}{2} \mathrm{k}\left(\mathrm{a}^{2}-\mathrm{x}^{2}\right)}{\frac{1}{2} \mathrm{kx}^{2}}=\frac{\mathrm{a}^{2}-\mathrm{x}^{2}}{\mathrm{x}^{2}}$ $\frac{\mathrm{K} . \mathrm{E}}{\mathrm{P} . \mathrm{E}} =\frac{\mathrm{a}^{2}-\left(\frac{\mathrm{a}}{2}\right)^{2}}{\left(\frac{\mathrm{a}}{2}\right)^{2}}=\frac{\frac{4 \mathrm{a}^{2}-\mathrm{a}^{2}}{4}}{\frac{\mathrm{a}^{2}}{4}}=\frac{\frac{3 \mathrm{a}^{2}}{4}}{\frac{\mathrm{a}^{2}}{4}}=\frac{3 \mathrm{a}^{2}}{\mathrm{a}^{2}}$ $\frac{\mathrm{K} . \mathrm{E}}{\mathrm{P} . \mathrm{E}} =\frac{3}{1}$
CG PET 2019
Oscillations
140369
Assertion: If the amplitude of a simple harmonic oscillator is doubled, its total energy becomes four times. Reason: The total energy is directly proportional to the square of the amplitude of vibration of the harmonic oscillator.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason in incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
A Total mechanical energy of a simple harmonic oscillator $\mathrm{E}=\frac{1}{2} \mathrm{ma}^{2} \omega^{2}$ $\mathrm{E} \propto \mathrm{a}^{2}$ $\therefore \quad \frac{E_{1}}{E_{2}}=\left(\frac{a_{1}}{a_{2}}\right)^{2}$ Given that, $a_{1}=a, a_{2}=2 a$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{\mathrm{a}}{2 \mathrm{a}}\right)^{2}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{1}{4}$ $4 \mathrm{E}_{1}=\mathrm{E}_{2}$
AIIMS-26.05.2018(E)
Oscillations
140370
The ratio of displacement to amplitude, when kinetic energy of a body executing SHM is thrice the potential energy
1 $\frac{3}{2}$
2 $\frac{4}{3}$
3 $\frac{1}{2}$
4 $\frac{2}{3}$
Explanation:
C Given that, K.E. $=3 \mathrm{U}$ We know that, K.E. in SHM- $\text { K.E. }=\frac{1}{2} m \omega^{2}\left(a^{2}-x^{2}\right)$ Potential energy in S.H.M. $\mathrm{U}=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{x}^{2}$ From equation (i), we get- $\frac{1}{2} m \omega^{2}\left(a^{2}-x^{2}\right)=3 \times \frac{1}{2} m \omega^{2} x^{2}$ $a^{2}-x^{2}=3 x^{2}$ $4 x^{2}=a^{2}$ $a=2 x$ $\frac{x}{a}=\frac{1}{2}$
140367
The angular retardation of a rotating flywheel is proportional to the angle through which it rotates. If its kinetic energy gets reduced by $\Delta \mathrm{E}$ while it rotates through at an angle $\theta$, then
A The condition of S.H.M is $-\mathrm{a} \propto \theta$ We know that, $\mathrm{f}_{\mathrm{r}}=-$ ma $\mathrm{f}_{\mathrm{r}} \propto \theta$ $\mathrm{f}_{\mathrm{r}}=\mathrm{k} \theta$ Law of conservation of energy- $\Delta \mathrm{W} =\Delta \mathrm{KE}$ $\Delta \mathrm{E} =\mathrm{f}_{\mathrm{r}} \cdot \theta$ $\Delta \mathrm{E} =\mathrm{k} \cdot \theta \cdot \theta \quad\left(\mathrm{f}_{\mathrm{r}}=\mathrm{k} \theta\right)$ $\Delta \mathrm{E} =\mathrm{k} \theta^{2}$ Hence, reduced kinetic energy $(\Delta \mathrm{E}) \propto \theta^{2}$
AP EAMCET (21.04.2019) Shift-II
Oscillations
140368
A particle starts oscillating in simple harmonic motion form its equilibrium position with time period $T$. The ratio of $K E$ and $P E$ of the particle at time $\mathrm{t}=\frac{\mathrm{T}}{12}$ is
1 $3: 1$
2 $1: 4$
3 $4: 1$
4 $2: 1$
Explanation:
A We know that, displacement equation in S.H.M is given by- $\mathrm{x}=\mathrm{a} \sin \omega \mathrm{t}, \quad \mathrm{t}=\frac{\mathrm{T}}{12}$ $x=a \sin \left(\omega \times \frac{T}{12}\right)$ $x=a \sin \left(\frac{2 \pi}{T} \times \frac{T}{12}\right)$ $x=a \sin 30^{\circ}$ $x=\frac{a}{2}$ Kinetic energy of the particle, $K E=\frac{1}{2} k\left(a^{2}-x^{2}\right)$ Potential energy $(P E)=\frac{1}{2} \mathrm{kx}^{2}$ $\therefore \quad \quad \frac{K . E}{P . E} =\frac{\frac{1}{2} \mathrm{k}\left(\mathrm{a}^{2}-\mathrm{x}^{2}\right)}{\frac{1}{2} \mathrm{kx}^{2}}=\frac{\mathrm{a}^{2}-\mathrm{x}^{2}}{\mathrm{x}^{2}}$ $\frac{\mathrm{K} . \mathrm{E}}{\mathrm{P} . \mathrm{E}} =\frac{\mathrm{a}^{2}-\left(\frac{\mathrm{a}}{2}\right)^{2}}{\left(\frac{\mathrm{a}}{2}\right)^{2}}=\frac{\frac{4 \mathrm{a}^{2}-\mathrm{a}^{2}}{4}}{\frac{\mathrm{a}^{2}}{4}}=\frac{\frac{3 \mathrm{a}^{2}}{4}}{\frac{\mathrm{a}^{2}}{4}}=\frac{3 \mathrm{a}^{2}}{\mathrm{a}^{2}}$ $\frac{\mathrm{K} . \mathrm{E}}{\mathrm{P} . \mathrm{E}} =\frac{3}{1}$
CG PET 2019
Oscillations
140369
Assertion: If the amplitude of a simple harmonic oscillator is doubled, its total energy becomes four times. Reason: The total energy is directly proportional to the square of the amplitude of vibration of the harmonic oscillator.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason in incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
A Total mechanical energy of a simple harmonic oscillator $\mathrm{E}=\frac{1}{2} \mathrm{ma}^{2} \omega^{2}$ $\mathrm{E} \propto \mathrm{a}^{2}$ $\therefore \quad \frac{E_{1}}{E_{2}}=\left(\frac{a_{1}}{a_{2}}\right)^{2}$ Given that, $a_{1}=a, a_{2}=2 a$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{\mathrm{a}}{2 \mathrm{a}}\right)^{2}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{1}{4}$ $4 \mathrm{E}_{1}=\mathrm{E}_{2}$
AIIMS-26.05.2018(E)
Oscillations
140370
The ratio of displacement to amplitude, when kinetic energy of a body executing SHM is thrice the potential energy
1 $\frac{3}{2}$
2 $\frac{4}{3}$
3 $\frac{1}{2}$
4 $\frac{2}{3}$
Explanation:
C Given that, K.E. $=3 \mathrm{U}$ We know that, K.E. in SHM- $\text { K.E. }=\frac{1}{2} m \omega^{2}\left(a^{2}-x^{2}\right)$ Potential energy in S.H.M. $\mathrm{U}=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{x}^{2}$ From equation (i), we get- $\frac{1}{2} m \omega^{2}\left(a^{2}-x^{2}\right)=3 \times \frac{1}{2} m \omega^{2} x^{2}$ $a^{2}-x^{2}=3 x^{2}$ $4 x^{2}=a^{2}$ $a=2 x$ $\frac{x}{a}=\frac{1}{2}$
140367
The angular retardation of a rotating flywheel is proportional to the angle through which it rotates. If its kinetic energy gets reduced by $\Delta \mathrm{E}$ while it rotates through at an angle $\theta$, then
A The condition of S.H.M is $-\mathrm{a} \propto \theta$ We know that, $\mathrm{f}_{\mathrm{r}}=-$ ma $\mathrm{f}_{\mathrm{r}} \propto \theta$ $\mathrm{f}_{\mathrm{r}}=\mathrm{k} \theta$ Law of conservation of energy- $\Delta \mathrm{W} =\Delta \mathrm{KE}$ $\Delta \mathrm{E} =\mathrm{f}_{\mathrm{r}} \cdot \theta$ $\Delta \mathrm{E} =\mathrm{k} \cdot \theta \cdot \theta \quad\left(\mathrm{f}_{\mathrm{r}}=\mathrm{k} \theta\right)$ $\Delta \mathrm{E} =\mathrm{k} \theta^{2}$ Hence, reduced kinetic energy $(\Delta \mathrm{E}) \propto \theta^{2}$
AP EAMCET (21.04.2019) Shift-II
Oscillations
140368
A particle starts oscillating in simple harmonic motion form its equilibrium position with time period $T$. The ratio of $K E$ and $P E$ of the particle at time $\mathrm{t}=\frac{\mathrm{T}}{12}$ is
1 $3: 1$
2 $1: 4$
3 $4: 1$
4 $2: 1$
Explanation:
A We know that, displacement equation in S.H.M is given by- $\mathrm{x}=\mathrm{a} \sin \omega \mathrm{t}, \quad \mathrm{t}=\frac{\mathrm{T}}{12}$ $x=a \sin \left(\omega \times \frac{T}{12}\right)$ $x=a \sin \left(\frac{2 \pi}{T} \times \frac{T}{12}\right)$ $x=a \sin 30^{\circ}$ $x=\frac{a}{2}$ Kinetic energy of the particle, $K E=\frac{1}{2} k\left(a^{2}-x^{2}\right)$ Potential energy $(P E)=\frac{1}{2} \mathrm{kx}^{2}$ $\therefore \quad \quad \frac{K . E}{P . E} =\frac{\frac{1}{2} \mathrm{k}\left(\mathrm{a}^{2}-\mathrm{x}^{2}\right)}{\frac{1}{2} \mathrm{kx}^{2}}=\frac{\mathrm{a}^{2}-\mathrm{x}^{2}}{\mathrm{x}^{2}}$ $\frac{\mathrm{K} . \mathrm{E}}{\mathrm{P} . \mathrm{E}} =\frac{\mathrm{a}^{2}-\left(\frac{\mathrm{a}}{2}\right)^{2}}{\left(\frac{\mathrm{a}}{2}\right)^{2}}=\frac{\frac{4 \mathrm{a}^{2}-\mathrm{a}^{2}}{4}}{\frac{\mathrm{a}^{2}}{4}}=\frac{\frac{3 \mathrm{a}^{2}}{4}}{\frac{\mathrm{a}^{2}}{4}}=\frac{3 \mathrm{a}^{2}}{\mathrm{a}^{2}}$ $\frac{\mathrm{K} . \mathrm{E}}{\mathrm{P} . \mathrm{E}} =\frac{3}{1}$
CG PET 2019
Oscillations
140369
Assertion: If the amplitude of a simple harmonic oscillator is doubled, its total energy becomes four times. Reason: The total energy is directly proportional to the square of the amplitude of vibration of the harmonic oscillator.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason in incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
A Total mechanical energy of a simple harmonic oscillator $\mathrm{E}=\frac{1}{2} \mathrm{ma}^{2} \omega^{2}$ $\mathrm{E} \propto \mathrm{a}^{2}$ $\therefore \quad \frac{E_{1}}{E_{2}}=\left(\frac{a_{1}}{a_{2}}\right)^{2}$ Given that, $a_{1}=a, a_{2}=2 a$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{\mathrm{a}}{2 \mathrm{a}}\right)^{2}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{1}{4}$ $4 \mathrm{E}_{1}=\mathrm{E}_{2}$
AIIMS-26.05.2018(E)
Oscillations
140370
The ratio of displacement to amplitude, when kinetic energy of a body executing SHM is thrice the potential energy
1 $\frac{3}{2}$
2 $\frac{4}{3}$
3 $\frac{1}{2}$
4 $\frac{2}{3}$
Explanation:
C Given that, K.E. $=3 \mathrm{U}$ We know that, K.E. in SHM- $\text { K.E. }=\frac{1}{2} m \omega^{2}\left(a^{2}-x^{2}\right)$ Potential energy in S.H.M. $\mathrm{U}=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{x}^{2}$ From equation (i), we get- $\frac{1}{2} m \omega^{2}\left(a^{2}-x^{2}\right)=3 \times \frac{1}{2} m \omega^{2} x^{2}$ $a^{2}-x^{2}=3 x^{2}$ $4 x^{2}=a^{2}$ $a=2 x$ $\frac{x}{a}=\frac{1}{2}$
