NEET Test Series from KOTA - 10 Papers In MS WORD
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Oscillations
140393
A particle of mass $4 \mathrm{~kg}$ is executing S.H.M . Its displacement is given by the equation $Y=8 \cos$ $[100 \mathrm{t}+\pi / 4] \mathrm{cm}$. Its maximum kinetic energy is
1 $128 \mathrm{~J}$
2 $64 \mathrm{~J}$
3 $16 \mathrm{~J}$
4 $32 \mathrm{~J}$
Explanation:
A Given, mass of particle (m) $=4 \mathrm{~kg}, \omega=100$ $\mathrm{rad} / \mathrm{s}$ For a displacement equation is given as, $y=8 \cos [100 t+\pi / 4]$ Now comparing the above equation with standard displacement equation, $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}+\theta)$ Where, $\omega=100 \mathrm{rad} / \mathrm{s}, \theta=\pi / 4$ $\mathrm{A}=8 \mathrm{~cm}=\frac{8}{100} \mathrm{~m}$ We know that, the maximum kinetic energy is given by $(\mathrm{KE})_{\max } =\frac{1}{2} \mathrm{~m}^{2} \mathrm{~A}^{2}$ $=\frac{1}{2} \times 4 \times 100 \times 100 \times \frac{8}{100} \times \frac{8}{100}$ $=2 \times 8 \times 8 \mathrm{~J}$ $=128 \mathrm{~J}$
TS EAMCET (Engg.)-2015
Oscillations
140394
An object of mass $0.2 \mathrm{~kg}$ executes SHM along $x$-axis with frequency of $\frac{25}{\pi}$ Hz. At the position $x=0.04 \mathrm{~m}$, the object has kinetic energy of $0.5 \mathrm{~J}$ and potential energy of $0.4 \mathrm{~J}$. The amplitude of oscillation is
1 $0.05 \mathrm{~m}$
2 $0.06 \mathrm{~m}$
3 $0.01 \mathrm{~m}$
4 None of these
Explanation:
B Given, mass of object $(\mathrm{m})=0.2 \mathrm{~kg}$, frequency $=\frac{25}{\pi}, \omega=2 \pi \mathrm{f}=2 \pi \times \frac{25}{\pi}=50 \mathrm{rad} / \mathrm{sec}$ Kinetic energy (K.E.) $=0.5 \mathrm{~J}$ Potential energy (P.E.) $=0.4 \mathrm{~J}$ $\therefore$ At $\mathrm{x}=0.01 \mathrm{~m}$ the total energy (TE) $\text { T.E. }=\text { K.E. + P.E. }$ $=0.5+0.4=0.9 \mathrm{~J}$ We know that - The total energy is given by - $\text { T.E. }=\frac{1}{2} \mathrm{~m}^{2} \mathrm{~A}^{2}$ $0.9=\frac{1}{2} \times 0.2 \times 50 \times 50 \times \mathrm{A}^{2}$ $0.9=250 \mathrm{~A}^{2}$ $\mathrm{~A}=0.06 \mathrm{~m}$
SRMJEEE - 2008
Oscillations
140395
The total energy of a body executing simple harmonic motion is $\mathbf{E}$. The kinetic energy when the displacement is $1 / 3$ of the amplitude
1 $\frac{\sqrt{3}}{8} \mathrm{E}$
2 $\frac{8}{\sqrt{3}} \mathrm{E}$
3 $\frac{8}{9} \mathrm{E}$
4 $\frac{3}{8} \mathrm{E}$
Explanation:
C Since, the total energy of a body executive simple harmonic motion (SHM) is E Total energy $(\mathrm{TE})=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}$ Potential energy $(E)=\frac{1}{2} m \omega^{2} A^{2}$ Kinetic energy $(K E)=\frac{1}{2} m \omega^{2}\left(A^{2}-y^{2}\right)$ We know that, T.E. $=$ K.E. + P.E. $\text { K.E. }=\mathrm{E}-\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{y}^{2}$ Hence, given the displacement is $\frac{1}{3}$ of the amplitude then, $\mathrm{y}=\frac{1}{3} \times \mathrm{A}=\frac{\mathrm{A}}{3}$ Putting the value of $y$ in equation (i) we get- $\text { K.E. }=\mathrm{E}-\frac{1}{2} \mathrm{~m} \omega^{2} \frac{\mathrm{A}^{2}}{9}$ $\text { K.E. }=\mathrm{E}-\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2} \times \frac{1}{9}$ $\text { K.E. }=\mathrm{E}-\frac{\mathrm{E}}{9}$ $\text { K.E. }=\frac{9 \mathrm{E}-\mathrm{E}}{9}=\frac{8}{9} \mathrm{E}$
SRMJEEE - 2014
Oscillations
140397
The ratio between kinetic and potential energies of a body executing simple harmonic motion, when it is at a distance of $\frac{1}{N}$ of its amplitude from the mean position is
1 $\mathrm{N}^{2}+1$
2 $\frac{1}{\mathrm{~N}^{2}}$
3 $\mathrm{N}^{2}$
4 $\mathrm{N}^{2}-1$
Explanation:
D Let the amplitude of oscillation of simple harmonic motion be $=\mathrm{A}$ Now, at mean position the distance of body be - $\mathrm{x}=\frac{\mathrm{A}}{\mathrm{N}}$ Kinetic energy $($ K.E $)=\frac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)$ $=\frac{1}{2} \mathrm{~m} \omega^{2}\left(\mathrm{~A}^{2}-\frac{\mathrm{A}^{2}}{\mathrm{~N}^{2}}\right)$ $=\frac{1}{2} \mathrm{~m} \omega^{2}\left(\frac{\mathrm{A}^{2} \mathrm{~N}^{2}-\mathrm{A}^{2}}{\mathrm{~N}^{2}}\right)$ $=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}\left(\frac{\mathrm{N}^{2}-1}{\mathrm{~N}^{2}}\right)$ Potential energy, $\mathrm{PE}=\frac{1}{2} \mathrm{kx}^{2}$ $\mathrm{PE}=\frac{1}{2} \mathrm{~m} \omega^{2} \frac{\mathrm{A}^{2}}{\mathrm{~N}^{2}} \quad\left[\because \mathrm{k}=\mathrm{m} \omega^{2}\right]$ From equation (i) and equation (ii), we get- $\frac{\text { K.E. }}{\text { P.E. }}=\frac{\frac{1}{2} \mathrm{~m}^{2} \mathrm{~A}^{2}\left(\frac{\mathrm{N}^{2}-1}{\mathrm{~N}^{2}}\right)}{\frac{1}{2} \mathrm{~m} \omega^{2} \frac{\mathrm{A}^{2}}{\mathrm{~N}^{2}}}$ $=\frac{\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}\left(\frac{\mathrm{N}^{2}-1}{\mathrm{~N}^{2}}\right)}{\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2} \frac{1}{\mathrm{~N}^{2}}}=\frac{\frac{\mathrm{N}^{2}-1}{\mathrm{~N}^{2}}}{\frac{1}{\mathrm{~N}^{2}}}$ $=\frac{\left(\mathrm{N}^{2}-1\right) \times \mathrm{N}^{2}}{\mathrm{~N}^{2}}=\mathrm{N}^{2}-1$
140393
A particle of mass $4 \mathrm{~kg}$ is executing S.H.M . Its displacement is given by the equation $Y=8 \cos$ $[100 \mathrm{t}+\pi / 4] \mathrm{cm}$. Its maximum kinetic energy is
1 $128 \mathrm{~J}$
2 $64 \mathrm{~J}$
3 $16 \mathrm{~J}$
4 $32 \mathrm{~J}$
Explanation:
A Given, mass of particle (m) $=4 \mathrm{~kg}, \omega=100$ $\mathrm{rad} / \mathrm{s}$ For a displacement equation is given as, $y=8 \cos [100 t+\pi / 4]$ Now comparing the above equation with standard displacement equation, $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}+\theta)$ Where, $\omega=100 \mathrm{rad} / \mathrm{s}, \theta=\pi / 4$ $\mathrm{A}=8 \mathrm{~cm}=\frac{8}{100} \mathrm{~m}$ We know that, the maximum kinetic energy is given by $(\mathrm{KE})_{\max } =\frac{1}{2} \mathrm{~m}^{2} \mathrm{~A}^{2}$ $=\frac{1}{2} \times 4 \times 100 \times 100 \times \frac{8}{100} \times \frac{8}{100}$ $=2 \times 8 \times 8 \mathrm{~J}$ $=128 \mathrm{~J}$
TS EAMCET (Engg.)-2015
Oscillations
140394
An object of mass $0.2 \mathrm{~kg}$ executes SHM along $x$-axis with frequency of $\frac{25}{\pi}$ Hz. At the position $x=0.04 \mathrm{~m}$, the object has kinetic energy of $0.5 \mathrm{~J}$ and potential energy of $0.4 \mathrm{~J}$. The amplitude of oscillation is
1 $0.05 \mathrm{~m}$
2 $0.06 \mathrm{~m}$
3 $0.01 \mathrm{~m}$
4 None of these
Explanation:
B Given, mass of object $(\mathrm{m})=0.2 \mathrm{~kg}$, frequency $=\frac{25}{\pi}, \omega=2 \pi \mathrm{f}=2 \pi \times \frac{25}{\pi}=50 \mathrm{rad} / \mathrm{sec}$ Kinetic energy (K.E.) $=0.5 \mathrm{~J}$ Potential energy (P.E.) $=0.4 \mathrm{~J}$ $\therefore$ At $\mathrm{x}=0.01 \mathrm{~m}$ the total energy (TE) $\text { T.E. }=\text { K.E. + P.E. }$ $=0.5+0.4=0.9 \mathrm{~J}$ We know that - The total energy is given by - $\text { T.E. }=\frac{1}{2} \mathrm{~m}^{2} \mathrm{~A}^{2}$ $0.9=\frac{1}{2} \times 0.2 \times 50 \times 50 \times \mathrm{A}^{2}$ $0.9=250 \mathrm{~A}^{2}$ $\mathrm{~A}=0.06 \mathrm{~m}$
SRMJEEE - 2008
Oscillations
140395
The total energy of a body executing simple harmonic motion is $\mathbf{E}$. The kinetic energy when the displacement is $1 / 3$ of the amplitude
1 $\frac{\sqrt{3}}{8} \mathrm{E}$
2 $\frac{8}{\sqrt{3}} \mathrm{E}$
3 $\frac{8}{9} \mathrm{E}$
4 $\frac{3}{8} \mathrm{E}$
Explanation:
C Since, the total energy of a body executive simple harmonic motion (SHM) is E Total energy $(\mathrm{TE})=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}$ Potential energy $(E)=\frac{1}{2} m \omega^{2} A^{2}$ Kinetic energy $(K E)=\frac{1}{2} m \omega^{2}\left(A^{2}-y^{2}\right)$ We know that, T.E. $=$ K.E. + P.E. $\text { K.E. }=\mathrm{E}-\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{y}^{2}$ Hence, given the displacement is $\frac{1}{3}$ of the amplitude then, $\mathrm{y}=\frac{1}{3} \times \mathrm{A}=\frac{\mathrm{A}}{3}$ Putting the value of $y$ in equation (i) we get- $\text { K.E. }=\mathrm{E}-\frac{1}{2} \mathrm{~m} \omega^{2} \frac{\mathrm{A}^{2}}{9}$ $\text { K.E. }=\mathrm{E}-\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2} \times \frac{1}{9}$ $\text { K.E. }=\mathrm{E}-\frac{\mathrm{E}}{9}$ $\text { K.E. }=\frac{9 \mathrm{E}-\mathrm{E}}{9}=\frac{8}{9} \mathrm{E}$
SRMJEEE - 2014
Oscillations
140397
The ratio between kinetic and potential energies of a body executing simple harmonic motion, when it is at a distance of $\frac{1}{N}$ of its amplitude from the mean position is
1 $\mathrm{N}^{2}+1$
2 $\frac{1}{\mathrm{~N}^{2}}$
3 $\mathrm{N}^{2}$
4 $\mathrm{N}^{2}-1$
Explanation:
D Let the amplitude of oscillation of simple harmonic motion be $=\mathrm{A}$ Now, at mean position the distance of body be - $\mathrm{x}=\frac{\mathrm{A}}{\mathrm{N}}$ Kinetic energy $($ K.E $)=\frac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)$ $=\frac{1}{2} \mathrm{~m} \omega^{2}\left(\mathrm{~A}^{2}-\frac{\mathrm{A}^{2}}{\mathrm{~N}^{2}}\right)$ $=\frac{1}{2} \mathrm{~m} \omega^{2}\left(\frac{\mathrm{A}^{2} \mathrm{~N}^{2}-\mathrm{A}^{2}}{\mathrm{~N}^{2}}\right)$ $=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}\left(\frac{\mathrm{N}^{2}-1}{\mathrm{~N}^{2}}\right)$ Potential energy, $\mathrm{PE}=\frac{1}{2} \mathrm{kx}^{2}$ $\mathrm{PE}=\frac{1}{2} \mathrm{~m} \omega^{2} \frac{\mathrm{A}^{2}}{\mathrm{~N}^{2}} \quad\left[\because \mathrm{k}=\mathrm{m} \omega^{2}\right]$ From equation (i) and equation (ii), we get- $\frac{\text { K.E. }}{\text { P.E. }}=\frac{\frac{1}{2} \mathrm{~m}^{2} \mathrm{~A}^{2}\left(\frac{\mathrm{N}^{2}-1}{\mathrm{~N}^{2}}\right)}{\frac{1}{2} \mathrm{~m} \omega^{2} \frac{\mathrm{A}^{2}}{\mathrm{~N}^{2}}}$ $=\frac{\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}\left(\frac{\mathrm{N}^{2}-1}{\mathrm{~N}^{2}}\right)}{\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2} \frac{1}{\mathrm{~N}^{2}}}=\frac{\frac{\mathrm{N}^{2}-1}{\mathrm{~N}^{2}}}{\frac{1}{\mathrm{~N}^{2}}}$ $=\frac{\left(\mathrm{N}^{2}-1\right) \times \mathrm{N}^{2}}{\mathrm{~N}^{2}}=\mathrm{N}^{2}-1$
140393
A particle of mass $4 \mathrm{~kg}$ is executing S.H.M . Its displacement is given by the equation $Y=8 \cos$ $[100 \mathrm{t}+\pi / 4] \mathrm{cm}$. Its maximum kinetic energy is
1 $128 \mathrm{~J}$
2 $64 \mathrm{~J}$
3 $16 \mathrm{~J}$
4 $32 \mathrm{~J}$
Explanation:
A Given, mass of particle (m) $=4 \mathrm{~kg}, \omega=100$ $\mathrm{rad} / \mathrm{s}$ For a displacement equation is given as, $y=8 \cos [100 t+\pi / 4]$ Now comparing the above equation with standard displacement equation, $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}+\theta)$ Where, $\omega=100 \mathrm{rad} / \mathrm{s}, \theta=\pi / 4$ $\mathrm{A}=8 \mathrm{~cm}=\frac{8}{100} \mathrm{~m}$ We know that, the maximum kinetic energy is given by $(\mathrm{KE})_{\max } =\frac{1}{2} \mathrm{~m}^{2} \mathrm{~A}^{2}$ $=\frac{1}{2} \times 4 \times 100 \times 100 \times \frac{8}{100} \times \frac{8}{100}$ $=2 \times 8 \times 8 \mathrm{~J}$ $=128 \mathrm{~J}$
TS EAMCET (Engg.)-2015
Oscillations
140394
An object of mass $0.2 \mathrm{~kg}$ executes SHM along $x$-axis with frequency of $\frac{25}{\pi}$ Hz. At the position $x=0.04 \mathrm{~m}$, the object has kinetic energy of $0.5 \mathrm{~J}$ and potential energy of $0.4 \mathrm{~J}$. The amplitude of oscillation is
1 $0.05 \mathrm{~m}$
2 $0.06 \mathrm{~m}$
3 $0.01 \mathrm{~m}$
4 None of these
Explanation:
B Given, mass of object $(\mathrm{m})=0.2 \mathrm{~kg}$, frequency $=\frac{25}{\pi}, \omega=2 \pi \mathrm{f}=2 \pi \times \frac{25}{\pi}=50 \mathrm{rad} / \mathrm{sec}$ Kinetic energy (K.E.) $=0.5 \mathrm{~J}$ Potential energy (P.E.) $=0.4 \mathrm{~J}$ $\therefore$ At $\mathrm{x}=0.01 \mathrm{~m}$ the total energy (TE) $\text { T.E. }=\text { K.E. + P.E. }$ $=0.5+0.4=0.9 \mathrm{~J}$ We know that - The total energy is given by - $\text { T.E. }=\frac{1}{2} \mathrm{~m}^{2} \mathrm{~A}^{2}$ $0.9=\frac{1}{2} \times 0.2 \times 50 \times 50 \times \mathrm{A}^{2}$ $0.9=250 \mathrm{~A}^{2}$ $\mathrm{~A}=0.06 \mathrm{~m}$
SRMJEEE - 2008
Oscillations
140395
The total energy of a body executing simple harmonic motion is $\mathbf{E}$. The kinetic energy when the displacement is $1 / 3$ of the amplitude
1 $\frac{\sqrt{3}}{8} \mathrm{E}$
2 $\frac{8}{\sqrt{3}} \mathrm{E}$
3 $\frac{8}{9} \mathrm{E}$
4 $\frac{3}{8} \mathrm{E}$
Explanation:
C Since, the total energy of a body executive simple harmonic motion (SHM) is E Total energy $(\mathrm{TE})=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}$ Potential energy $(E)=\frac{1}{2} m \omega^{2} A^{2}$ Kinetic energy $(K E)=\frac{1}{2} m \omega^{2}\left(A^{2}-y^{2}\right)$ We know that, T.E. $=$ K.E. + P.E. $\text { K.E. }=\mathrm{E}-\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{y}^{2}$ Hence, given the displacement is $\frac{1}{3}$ of the amplitude then, $\mathrm{y}=\frac{1}{3} \times \mathrm{A}=\frac{\mathrm{A}}{3}$ Putting the value of $y$ in equation (i) we get- $\text { K.E. }=\mathrm{E}-\frac{1}{2} \mathrm{~m} \omega^{2} \frac{\mathrm{A}^{2}}{9}$ $\text { K.E. }=\mathrm{E}-\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2} \times \frac{1}{9}$ $\text { K.E. }=\mathrm{E}-\frac{\mathrm{E}}{9}$ $\text { K.E. }=\frac{9 \mathrm{E}-\mathrm{E}}{9}=\frac{8}{9} \mathrm{E}$
SRMJEEE - 2014
Oscillations
140397
The ratio between kinetic and potential energies of a body executing simple harmonic motion, when it is at a distance of $\frac{1}{N}$ of its amplitude from the mean position is
1 $\mathrm{N}^{2}+1$
2 $\frac{1}{\mathrm{~N}^{2}}$
3 $\mathrm{N}^{2}$
4 $\mathrm{N}^{2}-1$
Explanation:
D Let the amplitude of oscillation of simple harmonic motion be $=\mathrm{A}$ Now, at mean position the distance of body be - $\mathrm{x}=\frac{\mathrm{A}}{\mathrm{N}}$ Kinetic energy $($ K.E $)=\frac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)$ $=\frac{1}{2} \mathrm{~m} \omega^{2}\left(\mathrm{~A}^{2}-\frac{\mathrm{A}^{2}}{\mathrm{~N}^{2}}\right)$ $=\frac{1}{2} \mathrm{~m} \omega^{2}\left(\frac{\mathrm{A}^{2} \mathrm{~N}^{2}-\mathrm{A}^{2}}{\mathrm{~N}^{2}}\right)$ $=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}\left(\frac{\mathrm{N}^{2}-1}{\mathrm{~N}^{2}}\right)$ Potential energy, $\mathrm{PE}=\frac{1}{2} \mathrm{kx}^{2}$ $\mathrm{PE}=\frac{1}{2} \mathrm{~m} \omega^{2} \frac{\mathrm{A}^{2}}{\mathrm{~N}^{2}} \quad\left[\because \mathrm{k}=\mathrm{m} \omega^{2}\right]$ From equation (i) and equation (ii), we get- $\frac{\text { K.E. }}{\text { P.E. }}=\frac{\frac{1}{2} \mathrm{~m}^{2} \mathrm{~A}^{2}\left(\frac{\mathrm{N}^{2}-1}{\mathrm{~N}^{2}}\right)}{\frac{1}{2} \mathrm{~m} \omega^{2} \frac{\mathrm{A}^{2}}{\mathrm{~N}^{2}}}$ $=\frac{\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}\left(\frac{\mathrm{N}^{2}-1}{\mathrm{~N}^{2}}\right)}{\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2} \frac{1}{\mathrm{~N}^{2}}}=\frac{\frac{\mathrm{N}^{2}-1}{\mathrm{~N}^{2}}}{\frac{1}{\mathrm{~N}^{2}}}$ $=\frac{\left(\mathrm{N}^{2}-1\right) \times \mathrm{N}^{2}}{\mathrm{~N}^{2}}=\mathrm{N}^{2}-1$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Oscillations
140393
A particle of mass $4 \mathrm{~kg}$ is executing S.H.M . Its displacement is given by the equation $Y=8 \cos$ $[100 \mathrm{t}+\pi / 4] \mathrm{cm}$. Its maximum kinetic energy is
1 $128 \mathrm{~J}$
2 $64 \mathrm{~J}$
3 $16 \mathrm{~J}$
4 $32 \mathrm{~J}$
Explanation:
A Given, mass of particle (m) $=4 \mathrm{~kg}, \omega=100$ $\mathrm{rad} / \mathrm{s}$ For a displacement equation is given as, $y=8 \cos [100 t+\pi / 4]$ Now comparing the above equation with standard displacement equation, $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}+\theta)$ Where, $\omega=100 \mathrm{rad} / \mathrm{s}, \theta=\pi / 4$ $\mathrm{A}=8 \mathrm{~cm}=\frac{8}{100} \mathrm{~m}$ We know that, the maximum kinetic energy is given by $(\mathrm{KE})_{\max } =\frac{1}{2} \mathrm{~m}^{2} \mathrm{~A}^{2}$ $=\frac{1}{2} \times 4 \times 100 \times 100 \times \frac{8}{100} \times \frac{8}{100}$ $=2 \times 8 \times 8 \mathrm{~J}$ $=128 \mathrm{~J}$
TS EAMCET (Engg.)-2015
Oscillations
140394
An object of mass $0.2 \mathrm{~kg}$ executes SHM along $x$-axis with frequency of $\frac{25}{\pi}$ Hz. At the position $x=0.04 \mathrm{~m}$, the object has kinetic energy of $0.5 \mathrm{~J}$ and potential energy of $0.4 \mathrm{~J}$. The amplitude of oscillation is
1 $0.05 \mathrm{~m}$
2 $0.06 \mathrm{~m}$
3 $0.01 \mathrm{~m}$
4 None of these
Explanation:
B Given, mass of object $(\mathrm{m})=0.2 \mathrm{~kg}$, frequency $=\frac{25}{\pi}, \omega=2 \pi \mathrm{f}=2 \pi \times \frac{25}{\pi}=50 \mathrm{rad} / \mathrm{sec}$ Kinetic energy (K.E.) $=0.5 \mathrm{~J}$ Potential energy (P.E.) $=0.4 \mathrm{~J}$ $\therefore$ At $\mathrm{x}=0.01 \mathrm{~m}$ the total energy (TE) $\text { T.E. }=\text { K.E. + P.E. }$ $=0.5+0.4=0.9 \mathrm{~J}$ We know that - The total energy is given by - $\text { T.E. }=\frac{1}{2} \mathrm{~m}^{2} \mathrm{~A}^{2}$ $0.9=\frac{1}{2} \times 0.2 \times 50 \times 50 \times \mathrm{A}^{2}$ $0.9=250 \mathrm{~A}^{2}$ $\mathrm{~A}=0.06 \mathrm{~m}$
SRMJEEE - 2008
Oscillations
140395
The total energy of a body executing simple harmonic motion is $\mathbf{E}$. The kinetic energy when the displacement is $1 / 3$ of the amplitude
1 $\frac{\sqrt{3}}{8} \mathrm{E}$
2 $\frac{8}{\sqrt{3}} \mathrm{E}$
3 $\frac{8}{9} \mathrm{E}$
4 $\frac{3}{8} \mathrm{E}$
Explanation:
C Since, the total energy of a body executive simple harmonic motion (SHM) is E Total energy $(\mathrm{TE})=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}$ Potential energy $(E)=\frac{1}{2} m \omega^{2} A^{2}$ Kinetic energy $(K E)=\frac{1}{2} m \omega^{2}\left(A^{2}-y^{2}\right)$ We know that, T.E. $=$ K.E. + P.E. $\text { K.E. }=\mathrm{E}-\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{y}^{2}$ Hence, given the displacement is $\frac{1}{3}$ of the amplitude then, $\mathrm{y}=\frac{1}{3} \times \mathrm{A}=\frac{\mathrm{A}}{3}$ Putting the value of $y$ in equation (i) we get- $\text { K.E. }=\mathrm{E}-\frac{1}{2} \mathrm{~m} \omega^{2} \frac{\mathrm{A}^{2}}{9}$ $\text { K.E. }=\mathrm{E}-\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2} \times \frac{1}{9}$ $\text { K.E. }=\mathrm{E}-\frac{\mathrm{E}}{9}$ $\text { K.E. }=\frac{9 \mathrm{E}-\mathrm{E}}{9}=\frac{8}{9} \mathrm{E}$
SRMJEEE - 2014
Oscillations
140397
The ratio between kinetic and potential energies of a body executing simple harmonic motion, when it is at a distance of $\frac{1}{N}$ of its amplitude from the mean position is
1 $\mathrm{N}^{2}+1$
2 $\frac{1}{\mathrm{~N}^{2}}$
3 $\mathrm{N}^{2}$
4 $\mathrm{N}^{2}-1$
Explanation:
D Let the amplitude of oscillation of simple harmonic motion be $=\mathrm{A}$ Now, at mean position the distance of body be - $\mathrm{x}=\frac{\mathrm{A}}{\mathrm{N}}$ Kinetic energy $($ K.E $)=\frac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)$ $=\frac{1}{2} \mathrm{~m} \omega^{2}\left(\mathrm{~A}^{2}-\frac{\mathrm{A}^{2}}{\mathrm{~N}^{2}}\right)$ $=\frac{1}{2} \mathrm{~m} \omega^{2}\left(\frac{\mathrm{A}^{2} \mathrm{~N}^{2}-\mathrm{A}^{2}}{\mathrm{~N}^{2}}\right)$ $=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}\left(\frac{\mathrm{N}^{2}-1}{\mathrm{~N}^{2}}\right)$ Potential energy, $\mathrm{PE}=\frac{1}{2} \mathrm{kx}^{2}$ $\mathrm{PE}=\frac{1}{2} \mathrm{~m} \omega^{2} \frac{\mathrm{A}^{2}}{\mathrm{~N}^{2}} \quad\left[\because \mathrm{k}=\mathrm{m} \omega^{2}\right]$ From equation (i) and equation (ii), we get- $\frac{\text { K.E. }}{\text { P.E. }}=\frac{\frac{1}{2} \mathrm{~m}^{2} \mathrm{~A}^{2}\left(\frac{\mathrm{N}^{2}-1}{\mathrm{~N}^{2}}\right)}{\frac{1}{2} \mathrm{~m} \omega^{2} \frac{\mathrm{A}^{2}}{\mathrm{~N}^{2}}}$ $=\frac{\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}\left(\frac{\mathrm{N}^{2}-1}{\mathrm{~N}^{2}}\right)}{\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2} \frac{1}{\mathrm{~N}^{2}}}=\frac{\frac{\mathrm{N}^{2}-1}{\mathrm{~N}^{2}}}{\frac{1}{\mathrm{~N}^{2}}}$ $=\frac{\left(\mathrm{N}^{2}-1\right) \times \mathrm{N}^{2}}{\mathrm{~N}^{2}}=\mathrm{N}^{2}-1$