140383
The total energy of the particle executing simple harmonic of amplitude $A$ is $100 \mathrm{~J}$. At a distance of $0.707 \mathrm{~A}$ from the mean position, its kinetic energy is
140384
A particle executes simple harmonic motion with a frequency $f$. The frequency with which its kinetic energy oscillates is:
1 $f / 2$
2 $\mathrm{f}$
3 $2 \mathrm{f}$
4 $4 \mathrm{f}$
Explanation:
C Given, frequency $=\mathrm{f}$ Displacement equation for a particle executing SHM $\mathrm{y}=\mathrm{a} \sin \omega \mathrm{t}$ Differentiating both side w.r.t time, we get- $\mathrm{v}=\frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{a} \cos (\omega \mathrm{t}) \cdot \omega$ $\mathrm{v}=\mathrm{a} \omega \cos \omega \mathrm{t}$ Kinetic energy $=\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \mathrm{ma}^{2} \omega^{2} \cos ^{2} \omega \mathrm{t}$ $\mathrm{KE}=\frac{1}{2} \mathrm{ma}^{2} \omega^{2}\left(\frac{\cos 2 \omega \mathrm{t}+1}{2}\right)$ $\because \cos 2 \theta=2 \cos ^{2} \theta-1$ $\cos ^{2} \theta=\frac{\cos 2 \theta+1}{2}$ $\mathrm{KE}=\frac{1}{4} \mathrm{ma}^{2} \omega^{2}(\cos 2 \omega \mathrm{t}+1)$ $\mathrm{KE}=\frac{1}{4} \mathrm{ma}^{2} \mathrm{f}^{2}(\cos 2 \mathrm{ft}+1) \quad\{\omega=\mathrm{f}\}$ $\mathrm{KE}=\frac{1}{4} \mathrm{ma}^{2} \mathrm{f}^{2}\left(\cos _{\mathrm{k}} \mathrm{t}+1\right) \quad\left(\right.$ Where, $\left.\mathrm{f}_{\mathrm{k}}=2 \mathrm{f}\right)$ Hence, frequency of oscillation of kinetic energy is $2 \mathrm{f}$.
JCECE-2006
Oscillations
140386
At a displacement from the equilibrium position, that is one- half the amplitude of oscillation, what fraction of the total energy of the oscillator is kinetic energy?
1 $\frac{1}{2}$
2 $\frac{1}{4}$
3 $\frac{1}{\sqrt{2}}$
4 $\frac{3}{4}$
Explanation:
D Given, displacement $(\mathrm{x})=\frac{\mathrm{a}}{2}$ We know that, potential energy of a simple harmonic oscillator $\mathrm{U}=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{a}^{2}$ And kinetic energy of a simple harmonic motion when displacement is $\mathrm{x}$ $=\frac{1}{2} \mathrm{~m} \omega^{2}\left(\mathrm{a}^{2}-\mathrm{x}^{2}\right)$ $=\frac{1}{2} \mathrm{~m} \omega^{2}\left(\mathrm{a}^{2}-\left(\frac{\mathrm{a}}{2}\right)^{2}\right)$ $=\frac{1}{2} \mathrm{~m} \omega^{2} \times \frac{3}{4} \mathrm{a}^{2}$ $=\frac{3}{4}\left(\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{a}^{2}\right)$ Therefore, fraction of the total energy at $\mathrm{x}=\frac{\frac{3}{4}\left(\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{a}^{2}\right)}{\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{a}^{2}}=\frac{3}{4}$
J and K-CET-2012
Oscillations
140387
The force constant of weightless spring is $16 \mathrm{~N} / \mathrm{m}$. A body of mass $1.0 \mathrm{~kg}$ suspended from it is pulled down through $5 \mathrm{~cm}$ and then released. The maximum kinetic energy of the system (spring + body) will be
1 $2 \times 10^{-2} \mathrm{~J}$
2 $4 \times 10^{-2} \mathrm{~J}$
3 $8 \times 10^{-2} \mathrm{~J}$
4 $16 \times 10^{-2} \mathrm{~J}$
Explanation:
A Given, Force constant $(\mathrm{k})=16 \mathrm{~N} / \mathrm{m}$ Mass of body $(\mathrm{m})=1.0 \mathrm{~kg}$ Elongation in the spring $(\mathrm{x})=5 \mathrm{~cm}$ Maximum K.E. of the system, $\text { K.E. }=\frac{1}{2} \mathrm{kx}^{2}$ $=\frac{1}{2} \times 16 \times\left(5 \times 10^{-2}\right)^{2}$ $=2 \times 10^{-2} \mathrm{~J}$
Manipal UGET-2011
Oscillations
140388
Potential energy in a spring when stretched by $2 \mathrm{~cm}$ is $U$. Its potential energy, when stretched by $10 \mathrm{~cm}$ is :
1 $\frac{U}{25}$
2 $\frac{U}{5}$
3 $25 \mathrm{U}$
4 $5 \mathrm{U}$
5 none of these
Explanation:
C We know that - Potential energy in a spring when stretched by $2 \mathrm{~cm}$. $\mathrm{U}=\frac{1}{2} \mathrm{kx}^{2}$ Where, $\mathrm{k}=$ spring constant $\mathrm{x}=\text { elongation of spring }$ $\mathrm{U}=\frac{1}{2} \mathrm{k}(2)^{2}$ $\mathrm{U}=\frac{1}{2} \mathrm{k} \times 4$ When spring is stretched by $10 \mathrm{~cm}$ then the potential energy will be, $\mathrm{U}^{\prime} =\frac{1}{2} \mathrm{k}(10)^{2}$ $\mathrm{U}^{\prime} =\frac{1}{2} \mathrm{k} \times 100$ Dividing equation (ii) by (i), we get - $\frac{\mathrm{U}^{\prime}}{\mathrm{U}}=\frac{\frac{1}{2} \mathrm{k} \times 100}{\frac{1}{2} \mathrm{k} \times 4}$ $\frac{\mathrm{U}^{\prime}}{\mathrm{U}}=25 \Rightarrow \mathrm{U}^{\prime}=25 \mathrm{U}$
140383
The total energy of the particle executing simple harmonic of amplitude $A$ is $100 \mathrm{~J}$. At a distance of $0.707 \mathrm{~A}$ from the mean position, its kinetic energy is
140384
A particle executes simple harmonic motion with a frequency $f$. The frequency with which its kinetic energy oscillates is:
1 $f / 2$
2 $\mathrm{f}$
3 $2 \mathrm{f}$
4 $4 \mathrm{f}$
Explanation:
C Given, frequency $=\mathrm{f}$ Displacement equation for a particle executing SHM $\mathrm{y}=\mathrm{a} \sin \omega \mathrm{t}$ Differentiating both side w.r.t time, we get- $\mathrm{v}=\frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{a} \cos (\omega \mathrm{t}) \cdot \omega$ $\mathrm{v}=\mathrm{a} \omega \cos \omega \mathrm{t}$ Kinetic energy $=\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \mathrm{ma}^{2} \omega^{2} \cos ^{2} \omega \mathrm{t}$ $\mathrm{KE}=\frac{1}{2} \mathrm{ma}^{2} \omega^{2}\left(\frac{\cos 2 \omega \mathrm{t}+1}{2}\right)$ $\because \cos 2 \theta=2 \cos ^{2} \theta-1$ $\cos ^{2} \theta=\frac{\cos 2 \theta+1}{2}$ $\mathrm{KE}=\frac{1}{4} \mathrm{ma}^{2} \omega^{2}(\cos 2 \omega \mathrm{t}+1)$ $\mathrm{KE}=\frac{1}{4} \mathrm{ma}^{2} \mathrm{f}^{2}(\cos 2 \mathrm{ft}+1) \quad\{\omega=\mathrm{f}\}$ $\mathrm{KE}=\frac{1}{4} \mathrm{ma}^{2} \mathrm{f}^{2}\left(\cos _{\mathrm{k}} \mathrm{t}+1\right) \quad\left(\right.$ Where, $\left.\mathrm{f}_{\mathrm{k}}=2 \mathrm{f}\right)$ Hence, frequency of oscillation of kinetic energy is $2 \mathrm{f}$.
JCECE-2006
Oscillations
140386
At a displacement from the equilibrium position, that is one- half the amplitude of oscillation, what fraction of the total energy of the oscillator is kinetic energy?
1 $\frac{1}{2}$
2 $\frac{1}{4}$
3 $\frac{1}{\sqrt{2}}$
4 $\frac{3}{4}$
Explanation:
D Given, displacement $(\mathrm{x})=\frac{\mathrm{a}}{2}$ We know that, potential energy of a simple harmonic oscillator $\mathrm{U}=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{a}^{2}$ And kinetic energy of a simple harmonic motion when displacement is $\mathrm{x}$ $=\frac{1}{2} \mathrm{~m} \omega^{2}\left(\mathrm{a}^{2}-\mathrm{x}^{2}\right)$ $=\frac{1}{2} \mathrm{~m} \omega^{2}\left(\mathrm{a}^{2}-\left(\frac{\mathrm{a}}{2}\right)^{2}\right)$ $=\frac{1}{2} \mathrm{~m} \omega^{2} \times \frac{3}{4} \mathrm{a}^{2}$ $=\frac{3}{4}\left(\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{a}^{2}\right)$ Therefore, fraction of the total energy at $\mathrm{x}=\frac{\frac{3}{4}\left(\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{a}^{2}\right)}{\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{a}^{2}}=\frac{3}{4}$
J and K-CET-2012
Oscillations
140387
The force constant of weightless spring is $16 \mathrm{~N} / \mathrm{m}$. A body of mass $1.0 \mathrm{~kg}$ suspended from it is pulled down through $5 \mathrm{~cm}$ and then released. The maximum kinetic energy of the system (spring + body) will be
1 $2 \times 10^{-2} \mathrm{~J}$
2 $4 \times 10^{-2} \mathrm{~J}$
3 $8 \times 10^{-2} \mathrm{~J}$
4 $16 \times 10^{-2} \mathrm{~J}$
Explanation:
A Given, Force constant $(\mathrm{k})=16 \mathrm{~N} / \mathrm{m}$ Mass of body $(\mathrm{m})=1.0 \mathrm{~kg}$ Elongation in the spring $(\mathrm{x})=5 \mathrm{~cm}$ Maximum K.E. of the system, $\text { K.E. }=\frac{1}{2} \mathrm{kx}^{2}$ $=\frac{1}{2} \times 16 \times\left(5 \times 10^{-2}\right)^{2}$ $=2 \times 10^{-2} \mathrm{~J}$
Manipal UGET-2011
Oscillations
140388
Potential energy in a spring when stretched by $2 \mathrm{~cm}$ is $U$. Its potential energy, when stretched by $10 \mathrm{~cm}$ is :
1 $\frac{U}{25}$
2 $\frac{U}{5}$
3 $25 \mathrm{U}$
4 $5 \mathrm{U}$
5 none of these
Explanation:
C We know that - Potential energy in a spring when stretched by $2 \mathrm{~cm}$. $\mathrm{U}=\frac{1}{2} \mathrm{kx}^{2}$ Where, $\mathrm{k}=$ spring constant $\mathrm{x}=\text { elongation of spring }$ $\mathrm{U}=\frac{1}{2} \mathrm{k}(2)^{2}$ $\mathrm{U}=\frac{1}{2} \mathrm{k} \times 4$ When spring is stretched by $10 \mathrm{~cm}$ then the potential energy will be, $\mathrm{U}^{\prime} =\frac{1}{2} \mathrm{k}(10)^{2}$ $\mathrm{U}^{\prime} =\frac{1}{2} \mathrm{k} \times 100$ Dividing equation (ii) by (i), we get - $\frac{\mathrm{U}^{\prime}}{\mathrm{U}}=\frac{\frac{1}{2} \mathrm{k} \times 100}{\frac{1}{2} \mathrm{k} \times 4}$ $\frac{\mathrm{U}^{\prime}}{\mathrm{U}}=25 \Rightarrow \mathrm{U}^{\prime}=25 \mathrm{U}$
140383
The total energy of the particle executing simple harmonic of amplitude $A$ is $100 \mathrm{~J}$. At a distance of $0.707 \mathrm{~A}$ from the mean position, its kinetic energy is
140384
A particle executes simple harmonic motion with a frequency $f$. The frequency with which its kinetic energy oscillates is:
1 $f / 2$
2 $\mathrm{f}$
3 $2 \mathrm{f}$
4 $4 \mathrm{f}$
Explanation:
C Given, frequency $=\mathrm{f}$ Displacement equation for a particle executing SHM $\mathrm{y}=\mathrm{a} \sin \omega \mathrm{t}$ Differentiating both side w.r.t time, we get- $\mathrm{v}=\frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{a} \cos (\omega \mathrm{t}) \cdot \omega$ $\mathrm{v}=\mathrm{a} \omega \cos \omega \mathrm{t}$ Kinetic energy $=\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \mathrm{ma}^{2} \omega^{2} \cos ^{2} \omega \mathrm{t}$ $\mathrm{KE}=\frac{1}{2} \mathrm{ma}^{2} \omega^{2}\left(\frac{\cos 2 \omega \mathrm{t}+1}{2}\right)$ $\because \cos 2 \theta=2 \cos ^{2} \theta-1$ $\cos ^{2} \theta=\frac{\cos 2 \theta+1}{2}$ $\mathrm{KE}=\frac{1}{4} \mathrm{ma}^{2} \omega^{2}(\cos 2 \omega \mathrm{t}+1)$ $\mathrm{KE}=\frac{1}{4} \mathrm{ma}^{2} \mathrm{f}^{2}(\cos 2 \mathrm{ft}+1) \quad\{\omega=\mathrm{f}\}$ $\mathrm{KE}=\frac{1}{4} \mathrm{ma}^{2} \mathrm{f}^{2}\left(\cos _{\mathrm{k}} \mathrm{t}+1\right) \quad\left(\right.$ Where, $\left.\mathrm{f}_{\mathrm{k}}=2 \mathrm{f}\right)$ Hence, frequency of oscillation of kinetic energy is $2 \mathrm{f}$.
JCECE-2006
Oscillations
140386
At a displacement from the equilibrium position, that is one- half the amplitude of oscillation, what fraction of the total energy of the oscillator is kinetic energy?
1 $\frac{1}{2}$
2 $\frac{1}{4}$
3 $\frac{1}{\sqrt{2}}$
4 $\frac{3}{4}$
Explanation:
D Given, displacement $(\mathrm{x})=\frac{\mathrm{a}}{2}$ We know that, potential energy of a simple harmonic oscillator $\mathrm{U}=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{a}^{2}$ And kinetic energy of a simple harmonic motion when displacement is $\mathrm{x}$ $=\frac{1}{2} \mathrm{~m} \omega^{2}\left(\mathrm{a}^{2}-\mathrm{x}^{2}\right)$ $=\frac{1}{2} \mathrm{~m} \omega^{2}\left(\mathrm{a}^{2}-\left(\frac{\mathrm{a}}{2}\right)^{2}\right)$ $=\frac{1}{2} \mathrm{~m} \omega^{2} \times \frac{3}{4} \mathrm{a}^{2}$ $=\frac{3}{4}\left(\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{a}^{2}\right)$ Therefore, fraction of the total energy at $\mathrm{x}=\frac{\frac{3}{4}\left(\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{a}^{2}\right)}{\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{a}^{2}}=\frac{3}{4}$
J and K-CET-2012
Oscillations
140387
The force constant of weightless spring is $16 \mathrm{~N} / \mathrm{m}$. A body of mass $1.0 \mathrm{~kg}$ suspended from it is pulled down through $5 \mathrm{~cm}$ and then released. The maximum kinetic energy of the system (spring + body) will be
1 $2 \times 10^{-2} \mathrm{~J}$
2 $4 \times 10^{-2} \mathrm{~J}$
3 $8 \times 10^{-2} \mathrm{~J}$
4 $16 \times 10^{-2} \mathrm{~J}$
Explanation:
A Given, Force constant $(\mathrm{k})=16 \mathrm{~N} / \mathrm{m}$ Mass of body $(\mathrm{m})=1.0 \mathrm{~kg}$ Elongation in the spring $(\mathrm{x})=5 \mathrm{~cm}$ Maximum K.E. of the system, $\text { K.E. }=\frac{1}{2} \mathrm{kx}^{2}$ $=\frac{1}{2} \times 16 \times\left(5 \times 10^{-2}\right)^{2}$ $=2 \times 10^{-2} \mathrm{~J}$
Manipal UGET-2011
Oscillations
140388
Potential energy in a spring when stretched by $2 \mathrm{~cm}$ is $U$. Its potential energy, when stretched by $10 \mathrm{~cm}$ is :
1 $\frac{U}{25}$
2 $\frac{U}{5}$
3 $25 \mathrm{U}$
4 $5 \mathrm{U}$
5 none of these
Explanation:
C We know that - Potential energy in a spring when stretched by $2 \mathrm{~cm}$. $\mathrm{U}=\frac{1}{2} \mathrm{kx}^{2}$ Where, $\mathrm{k}=$ spring constant $\mathrm{x}=\text { elongation of spring }$ $\mathrm{U}=\frac{1}{2} \mathrm{k}(2)^{2}$ $\mathrm{U}=\frac{1}{2} \mathrm{k} \times 4$ When spring is stretched by $10 \mathrm{~cm}$ then the potential energy will be, $\mathrm{U}^{\prime} =\frac{1}{2} \mathrm{k}(10)^{2}$ $\mathrm{U}^{\prime} =\frac{1}{2} \mathrm{k} \times 100$ Dividing equation (ii) by (i), we get - $\frac{\mathrm{U}^{\prime}}{\mathrm{U}}=\frac{\frac{1}{2} \mathrm{k} \times 100}{\frac{1}{2} \mathrm{k} \times 4}$ $\frac{\mathrm{U}^{\prime}}{\mathrm{U}}=25 \Rightarrow \mathrm{U}^{\prime}=25 \mathrm{U}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Oscillations
140383
The total energy of the particle executing simple harmonic of amplitude $A$ is $100 \mathrm{~J}$. At a distance of $0.707 \mathrm{~A}$ from the mean position, its kinetic energy is
140384
A particle executes simple harmonic motion with a frequency $f$. The frequency with which its kinetic energy oscillates is:
1 $f / 2$
2 $\mathrm{f}$
3 $2 \mathrm{f}$
4 $4 \mathrm{f}$
Explanation:
C Given, frequency $=\mathrm{f}$ Displacement equation for a particle executing SHM $\mathrm{y}=\mathrm{a} \sin \omega \mathrm{t}$ Differentiating both side w.r.t time, we get- $\mathrm{v}=\frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{a} \cos (\omega \mathrm{t}) \cdot \omega$ $\mathrm{v}=\mathrm{a} \omega \cos \omega \mathrm{t}$ Kinetic energy $=\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \mathrm{ma}^{2} \omega^{2} \cos ^{2} \omega \mathrm{t}$ $\mathrm{KE}=\frac{1}{2} \mathrm{ma}^{2} \omega^{2}\left(\frac{\cos 2 \omega \mathrm{t}+1}{2}\right)$ $\because \cos 2 \theta=2 \cos ^{2} \theta-1$ $\cos ^{2} \theta=\frac{\cos 2 \theta+1}{2}$ $\mathrm{KE}=\frac{1}{4} \mathrm{ma}^{2} \omega^{2}(\cos 2 \omega \mathrm{t}+1)$ $\mathrm{KE}=\frac{1}{4} \mathrm{ma}^{2} \mathrm{f}^{2}(\cos 2 \mathrm{ft}+1) \quad\{\omega=\mathrm{f}\}$ $\mathrm{KE}=\frac{1}{4} \mathrm{ma}^{2} \mathrm{f}^{2}\left(\cos _{\mathrm{k}} \mathrm{t}+1\right) \quad\left(\right.$ Where, $\left.\mathrm{f}_{\mathrm{k}}=2 \mathrm{f}\right)$ Hence, frequency of oscillation of kinetic energy is $2 \mathrm{f}$.
JCECE-2006
Oscillations
140386
At a displacement from the equilibrium position, that is one- half the amplitude of oscillation, what fraction of the total energy of the oscillator is kinetic energy?
1 $\frac{1}{2}$
2 $\frac{1}{4}$
3 $\frac{1}{\sqrt{2}}$
4 $\frac{3}{4}$
Explanation:
D Given, displacement $(\mathrm{x})=\frac{\mathrm{a}}{2}$ We know that, potential energy of a simple harmonic oscillator $\mathrm{U}=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{a}^{2}$ And kinetic energy of a simple harmonic motion when displacement is $\mathrm{x}$ $=\frac{1}{2} \mathrm{~m} \omega^{2}\left(\mathrm{a}^{2}-\mathrm{x}^{2}\right)$ $=\frac{1}{2} \mathrm{~m} \omega^{2}\left(\mathrm{a}^{2}-\left(\frac{\mathrm{a}}{2}\right)^{2}\right)$ $=\frac{1}{2} \mathrm{~m} \omega^{2} \times \frac{3}{4} \mathrm{a}^{2}$ $=\frac{3}{4}\left(\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{a}^{2}\right)$ Therefore, fraction of the total energy at $\mathrm{x}=\frac{\frac{3}{4}\left(\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{a}^{2}\right)}{\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{a}^{2}}=\frac{3}{4}$
J and K-CET-2012
Oscillations
140387
The force constant of weightless spring is $16 \mathrm{~N} / \mathrm{m}$. A body of mass $1.0 \mathrm{~kg}$ suspended from it is pulled down through $5 \mathrm{~cm}$ and then released. The maximum kinetic energy of the system (spring + body) will be
1 $2 \times 10^{-2} \mathrm{~J}$
2 $4 \times 10^{-2} \mathrm{~J}$
3 $8 \times 10^{-2} \mathrm{~J}$
4 $16 \times 10^{-2} \mathrm{~J}$
Explanation:
A Given, Force constant $(\mathrm{k})=16 \mathrm{~N} / \mathrm{m}$ Mass of body $(\mathrm{m})=1.0 \mathrm{~kg}$ Elongation in the spring $(\mathrm{x})=5 \mathrm{~cm}$ Maximum K.E. of the system, $\text { K.E. }=\frac{1}{2} \mathrm{kx}^{2}$ $=\frac{1}{2} \times 16 \times\left(5 \times 10^{-2}\right)^{2}$ $=2 \times 10^{-2} \mathrm{~J}$
Manipal UGET-2011
Oscillations
140388
Potential energy in a spring when stretched by $2 \mathrm{~cm}$ is $U$. Its potential energy, when stretched by $10 \mathrm{~cm}$ is :
1 $\frac{U}{25}$
2 $\frac{U}{5}$
3 $25 \mathrm{U}$
4 $5 \mathrm{U}$
5 none of these
Explanation:
C We know that - Potential energy in a spring when stretched by $2 \mathrm{~cm}$. $\mathrm{U}=\frac{1}{2} \mathrm{kx}^{2}$ Where, $\mathrm{k}=$ spring constant $\mathrm{x}=\text { elongation of spring }$ $\mathrm{U}=\frac{1}{2} \mathrm{k}(2)^{2}$ $\mathrm{U}=\frac{1}{2} \mathrm{k} \times 4$ When spring is stretched by $10 \mathrm{~cm}$ then the potential energy will be, $\mathrm{U}^{\prime} =\frac{1}{2} \mathrm{k}(10)^{2}$ $\mathrm{U}^{\prime} =\frac{1}{2} \mathrm{k} \times 100$ Dividing equation (ii) by (i), we get - $\frac{\mathrm{U}^{\prime}}{\mathrm{U}}=\frac{\frac{1}{2} \mathrm{k} \times 100}{\frac{1}{2} \mathrm{k} \times 4}$ $\frac{\mathrm{U}^{\prime}}{\mathrm{U}}=25 \Rightarrow \mathrm{U}^{\prime}=25 \mathrm{U}$
140383
The total energy of the particle executing simple harmonic of amplitude $A$ is $100 \mathrm{~J}$. At a distance of $0.707 \mathrm{~A}$ from the mean position, its kinetic energy is
140384
A particle executes simple harmonic motion with a frequency $f$. The frequency with which its kinetic energy oscillates is:
1 $f / 2$
2 $\mathrm{f}$
3 $2 \mathrm{f}$
4 $4 \mathrm{f}$
Explanation:
C Given, frequency $=\mathrm{f}$ Displacement equation for a particle executing SHM $\mathrm{y}=\mathrm{a} \sin \omega \mathrm{t}$ Differentiating both side w.r.t time, we get- $\mathrm{v}=\frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{a} \cos (\omega \mathrm{t}) \cdot \omega$ $\mathrm{v}=\mathrm{a} \omega \cos \omega \mathrm{t}$ Kinetic energy $=\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \mathrm{ma}^{2} \omega^{2} \cos ^{2} \omega \mathrm{t}$ $\mathrm{KE}=\frac{1}{2} \mathrm{ma}^{2} \omega^{2}\left(\frac{\cos 2 \omega \mathrm{t}+1}{2}\right)$ $\because \cos 2 \theta=2 \cos ^{2} \theta-1$ $\cos ^{2} \theta=\frac{\cos 2 \theta+1}{2}$ $\mathrm{KE}=\frac{1}{4} \mathrm{ma}^{2} \omega^{2}(\cos 2 \omega \mathrm{t}+1)$ $\mathrm{KE}=\frac{1}{4} \mathrm{ma}^{2} \mathrm{f}^{2}(\cos 2 \mathrm{ft}+1) \quad\{\omega=\mathrm{f}\}$ $\mathrm{KE}=\frac{1}{4} \mathrm{ma}^{2} \mathrm{f}^{2}\left(\cos _{\mathrm{k}} \mathrm{t}+1\right) \quad\left(\right.$ Where, $\left.\mathrm{f}_{\mathrm{k}}=2 \mathrm{f}\right)$ Hence, frequency of oscillation of kinetic energy is $2 \mathrm{f}$.
JCECE-2006
Oscillations
140386
At a displacement from the equilibrium position, that is one- half the amplitude of oscillation, what fraction of the total energy of the oscillator is kinetic energy?
1 $\frac{1}{2}$
2 $\frac{1}{4}$
3 $\frac{1}{\sqrt{2}}$
4 $\frac{3}{4}$
Explanation:
D Given, displacement $(\mathrm{x})=\frac{\mathrm{a}}{2}$ We know that, potential energy of a simple harmonic oscillator $\mathrm{U}=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{a}^{2}$ And kinetic energy of a simple harmonic motion when displacement is $\mathrm{x}$ $=\frac{1}{2} \mathrm{~m} \omega^{2}\left(\mathrm{a}^{2}-\mathrm{x}^{2}\right)$ $=\frac{1}{2} \mathrm{~m} \omega^{2}\left(\mathrm{a}^{2}-\left(\frac{\mathrm{a}}{2}\right)^{2}\right)$ $=\frac{1}{2} \mathrm{~m} \omega^{2} \times \frac{3}{4} \mathrm{a}^{2}$ $=\frac{3}{4}\left(\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{a}^{2}\right)$ Therefore, fraction of the total energy at $\mathrm{x}=\frac{\frac{3}{4}\left(\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{a}^{2}\right)}{\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{a}^{2}}=\frac{3}{4}$
J and K-CET-2012
Oscillations
140387
The force constant of weightless spring is $16 \mathrm{~N} / \mathrm{m}$. A body of mass $1.0 \mathrm{~kg}$ suspended from it is pulled down through $5 \mathrm{~cm}$ and then released. The maximum kinetic energy of the system (spring + body) will be
1 $2 \times 10^{-2} \mathrm{~J}$
2 $4 \times 10^{-2} \mathrm{~J}$
3 $8 \times 10^{-2} \mathrm{~J}$
4 $16 \times 10^{-2} \mathrm{~J}$
Explanation:
A Given, Force constant $(\mathrm{k})=16 \mathrm{~N} / \mathrm{m}$ Mass of body $(\mathrm{m})=1.0 \mathrm{~kg}$ Elongation in the spring $(\mathrm{x})=5 \mathrm{~cm}$ Maximum K.E. of the system, $\text { K.E. }=\frac{1}{2} \mathrm{kx}^{2}$ $=\frac{1}{2} \times 16 \times\left(5 \times 10^{-2}\right)^{2}$ $=2 \times 10^{-2} \mathrm{~J}$
Manipal UGET-2011
Oscillations
140388
Potential energy in a spring when stretched by $2 \mathrm{~cm}$ is $U$. Its potential energy, when stretched by $10 \mathrm{~cm}$ is :
1 $\frac{U}{25}$
2 $\frac{U}{5}$
3 $25 \mathrm{U}$
4 $5 \mathrm{U}$
5 none of these
Explanation:
C We know that - Potential energy in a spring when stretched by $2 \mathrm{~cm}$. $\mathrm{U}=\frac{1}{2} \mathrm{kx}^{2}$ Where, $\mathrm{k}=$ spring constant $\mathrm{x}=\text { elongation of spring }$ $\mathrm{U}=\frac{1}{2} \mathrm{k}(2)^{2}$ $\mathrm{U}=\frac{1}{2} \mathrm{k} \times 4$ When spring is stretched by $10 \mathrm{~cm}$ then the potential energy will be, $\mathrm{U}^{\prime} =\frac{1}{2} \mathrm{k}(10)^{2}$ $\mathrm{U}^{\prime} =\frac{1}{2} \mathrm{k} \times 100$ Dividing equation (ii) by (i), we get - $\frac{\mathrm{U}^{\prime}}{\mathrm{U}}=\frac{\frac{1}{2} \mathrm{k} \times 100}{\frac{1}{2} \mathrm{k} \times 4}$ $\frac{\mathrm{U}^{\prime}}{\mathrm{U}}=25 \Rightarrow \mathrm{U}^{\prime}=25 \mathrm{U}$