364094
The potential energy of a particle \(\left(U_{x}\right)\) executing SHM is given by
1 \(U_{x}=k_{1} x+k_{2} x^{2}+k_{3} x^{3}\)
2 \(U_{x}=\dfrac{k}{2}(x-a)^{2}\)
3 \(U_{x}=\) constant
4 \(U_{x}=A e^{-b x}\)
Explanation:
Potential energy of body in SHM at an instant, \(U=\dfrac{1}{2} k y^{2}\) where the displacement, \(y=(x-a)\)
PHXI14:OSCILLATIONS
364095
A particle of mass \(m\) moves in a one-dimensional potential energy \(U(x)=a x^{2}-b x^{4}\), where \(a\) and \(b\) are positive constants. The angular frequency of small oscillations about the minima of the potential energy is equal to
1 \(2 \sqrt{\dfrac{a}{m}}\)
2 \(\pi \sqrt{\dfrac{a}{2 b}}\)
3 \(\sqrt{\dfrac{a}{2 m}}\)
4 \(\sqrt{\dfrac{2 a}{m}}\)
Explanation:
\(U(x)=-a x^{2}+b x^{4}\) \(\therefore F=-\dfrac{d U}{d x}=-2 a x+4 b x^{3}\) At mean position \(F=0\) \(\begin{gathered}\therefore 2 a x=4 b x^{3} \\\text { or } x^{2}=\dfrac{2 a}{4 b}=\dfrac{a}{2 b} \Rightarrow x= \pm \sqrt{a / 2 b} \\\text { ' } \mathrm{U} \text { ' is minimum at } x=-\sqrt{\dfrac{a}{2 b}}\end{gathered}\) Now, \(F=m \dfrac{d^{2} x}{d t^{2}}\) \(\begin{aligned}& \therefore \quad m \dfrac{d^{2} x}{d t^{2}}=-2 a x+4 b x^{3} \\& \Rightarrow \dfrac{d^{2} x}{d t^{2}}=-\dfrac{2 a x}{m}\left(1-\dfrac{2 b}{a} x^{2}\right)\end{aligned}\) For small ' \(x\) ', \(\begin{aligned}& \dfrac{d^{2} x}{d t^{2}}=-\left(\dfrac{2 a}{m}\right) x \\& \therefore \quad \omega=\sqrt{2 a / m}\end{aligned}\)
PHXI14:OSCILLATIONS
364096
A body executes simple harmonic motion. The potential energy (PE), kinetic energy (KE) and total energy (TE) are measured as a function of displacement \(x\). Which of the following statement is true?
1 \(PE\) is maximum when \(x=0\)
2 \(TE\) is zero when \(x=0\)
3 \(KE\) is maximum when \(x\) is maximum
4 \(KE\) is maximum when \(x=0\)
Explanation:
\(P E=\dfrac{1}{2} m \omega^{2} x^{2}\) \(KE = \frac{1}{2}m{\omega ^2}\left( {{a^2} - {x^2}} \right)\) \(TE = PE + KE = \frac{1}{2}m{\omega ^2}{a^2}\) Since \(PE\) is maximum at \(x=a\) and \(KE\) is maximum at \(x=0\), therefore \(TE\) remains constant throughout the motion.
PHXI14:OSCILLATIONS
364097
The displacement of a particle executing SHM is given by \(y=5 \sin \left(4 t+\dfrac{\pi}{3}\right)\). If \(T\) is the time period and mass of the particle is \(2\;g\), the kinetic energy of the particle when \(t=\dfrac{T}{4}\) is given by
1 \(0.5\;J\)
2 \(0.4\;J\)
3 \(0.3\;J\)
4 \(3\;J\)
Explanation:
The displacement of particle, executing SHM, \(\begin{aligned}& y=5 \sin \left(4 t+\dfrac{\pi}{3}\right) \\& v=20 \cos \left(4 t+\dfrac{\pi}{3}\right)\end{aligned}\) Velocity at \(t=\left(\dfrac{T}{4}\right)\) \(\begin{aligned}& v=20 \cos \left(4 \times \dfrac{T}{4}+\dfrac{\pi}{3}\right) \\& \Rightarrow v=20 \cos \left(T+\dfrac{\pi}{3}\right)\end{aligned}\) Comparing the given equation with standard equation of SHM \(y=a \sin (\omega t+\phi)\), we get \(\omega=4\). \(\text { As } T=\dfrac{2 \pi}{\omega} \Rightarrow T=\dfrac{2 \pi}{4} \Rightarrow T=\dfrac{\pi}{2}\) Now, putting value of \(T\) in Eq. (2), we get \(\begin{aligned}v & =20 \cos \left(\dfrac{\pi}{2}+\dfrac{\pi}{3}\right)=-20 \sin \dfrac{\pi}{3} \\& =-20 \times \dfrac{\sqrt{3}}{2}=-10 \times \sqrt{3}\end{aligned}\) The kinetic energy of particle, \(KE = \frac{1}{2}m{v^2}\) \(\because m = 2g = 2 \times {10^{ - 3}}\;kg\) \( = \frac{1}{2} \times 2 \times {10^{ - 3}} \times {( - 10\sqrt 3 )^2}\) \( = {10^{ - 3}} \times 100 \times 3 = 3 \times {10^{ - 1}} \Rightarrow K.E = 0.3\,J.\)
PHXI14:OSCILLATIONS
364098
The kinetic energy of SHM is \(1 / n\) time its potential energy. If the amplitude of the SHM be \(A\), then what is the displacement of the particle?
1 \(\dfrac{A}{n}\)
2 \(\sqrt{\dfrac{n+1}{n}} A\)
3 \(\sqrt{\dfrac{n}{n+1}} A\)
4 \(n A\)
Explanation:
\(\begin{aligned}& K E=\dfrac{1}{n} P E \\& \dfrac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)=\dfrac{1}{n}\left(\dfrac{1}{2} m \omega^{2} x^{2}\right)\end{aligned}\) \(\Rightarrow x=\sqrt{\dfrac{n}{n+1}} A\)
364094
The potential energy of a particle \(\left(U_{x}\right)\) executing SHM is given by
1 \(U_{x}=k_{1} x+k_{2} x^{2}+k_{3} x^{3}\)
2 \(U_{x}=\dfrac{k}{2}(x-a)^{2}\)
3 \(U_{x}=\) constant
4 \(U_{x}=A e^{-b x}\)
Explanation:
Potential energy of body in SHM at an instant, \(U=\dfrac{1}{2} k y^{2}\) where the displacement, \(y=(x-a)\)
PHXI14:OSCILLATIONS
364095
A particle of mass \(m\) moves in a one-dimensional potential energy \(U(x)=a x^{2}-b x^{4}\), where \(a\) and \(b\) are positive constants. The angular frequency of small oscillations about the minima of the potential energy is equal to
1 \(2 \sqrt{\dfrac{a}{m}}\)
2 \(\pi \sqrt{\dfrac{a}{2 b}}\)
3 \(\sqrt{\dfrac{a}{2 m}}\)
4 \(\sqrt{\dfrac{2 a}{m}}\)
Explanation:
\(U(x)=-a x^{2}+b x^{4}\) \(\therefore F=-\dfrac{d U}{d x}=-2 a x+4 b x^{3}\) At mean position \(F=0\) \(\begin{gathered}\therefore 2 a x=4 b x^{3} \\\text { or } x^{2}=\dfrac{2 a}{4 b}=\dfrac{a}{2 b} \Rightarrow x= \pm \sqrt{a / 2 b} \\\text { ' } \mathrm{U} \text { ' is minimum at } x=-\sqrt{\dfrac{a}{2 b}}\end{gathered}\) Now, \(F=m \dfrac{d^{2} x}{d t^{2}}\) \(\begin{aligned}& \therefore \quad m \dfrac{d^{2} x}{d t^{2}}=-2 a x+4 b x^{3} \\& \Rightarrow \dfrac{d^{2} x}{d t^{2}}=-\dfrac{2 a x}{m}\left(1-\dfrac{2 b}{a} x^{2}\right)\end{aligned}\) For small ' \(x\) ', \(\begin{aligned}& \dfrac{d^{2} x}{d t^{2}}=-\left(\dfrac{2 a}{m}\right) x \\& \therefore \quad \omega=\sqrt{2 a / m}\end{aligned}\)
PHXI14:OSCILLATIONS
364096
A body executes simple harmonic motion. The potential energy (PE), kinetic energy (KE) and total energy (TE) are measured as a function of displacement \(x\). Which of the following statement is true?
1 \(PE\) is maximum when \(x=0\)
2 \(TE\) is zero when \(x=0\)
3 \(KE\) is maximum when \(x\) is maximum
4 \(KE\) is maximum when \(x=0\)
Explanation:
\(P E=\dfrac{1}{2} m \omega^{2} x^{2}\) \(KE = \frac{1}{2}m{\omega ^2}\left( {{a^2} - {x^2}} \right)\) \(TE = PE + KE = \frac{1}{2}m{\omega ^2}{a^2}\) Since \(PE\) is maximum at \(x=a\) and \(KE\) is maximum at \(x=0\), therefore \(TE\) remains constant throughout the motion.
PHXI14:OSCILLATIONS
364097
The displacement of a particle executing SHM is given by \(y=5 \sin \left(4 t+\dfrac{\pi}{3}\right)\). If \(T\) is the time period and mass of the particle is \(2\;g\), the kinetic energy of the particle when \(t=\dfrac{T}{4}\) is given by
1 \(0.5\;J\)
2 \(0.4\;J\)
3 \(0.3\;J\)
4 \(3\;J\)
Explanation:
The displacement of particle, executing SHM, \(\begin{aligned}& y=5 \sin \left(4 t+\dfrac{\pi}{3}\right) \\& v=20 \cos \left(4 t+\dfrac{\pi}{3}\right)\end{aligned}\) Velocity at \(t=\left(\dfrac{T}{4}\right)\) \(\begin{aligned}& v=20 \cos \left(4 \times \dfrac{T}{4}+\dfrac{\pi}{3}\right) \\& \Rightarrow v=20 \cos \left(T+\dfrac{\pi}{3}\right)\end{aligned}\) Comparing the given equation with standard equation of SHM \(y=a \sin (\omega t+\phi)\), we get \(\omega=4\). \(\text { As } T=\dfrac{2 \pi}{\omega} \Rightarrow T=\dfrac{2 \pi}{4} \Rightarrow T=\dfrac{\pi}{2}\) Now, putting value of \(T\) in Eq. (2), we get \(\begin{aligned}v & =20 \cos \left(\dfrac{\pi}{2}+\dfrac{\pi}{3}\right)=-20 \sin \dfrac{\pi}{3} \\& =-20 \times \dfrac{\sqrt{3}}{2}=-10 \times \sqrt{3}\end{aligned}\) The kinetic energy of particle, \(KE = \frac{1}{2}m{v^2}\) \(\because m = 2g = 2 \times {10^{ - 3}}\;kg\) \( = \frac{1}{2} \times 2 \times {10^{ - 3}} \times {( - 10\sqrt 3 )^2}\) \( = {10^{ - 3}} \times 100 \times 3 = 3 \times {10^{ - 1}} \Rightarrow K.E = 0.3\,J.\)
PHXI14:OSCILLATIONS
364098
The kinetic energy of SHM is \(1 / n\) time its potential energy. If the amplitude of the SHM be \(A\), then what is the displacement of the particle?
1 \(\dfrac{A}{n}\)
2 \(\sqrt{\dfrac{n+1}{n}} A\)
3 \(\sqrt{\dfrac{n}{n+1}} A\)
4 \(n A\)
Explanation:
\(\begin{aligned}& K E=\dfrac{1}{n} P E \\& \dfrac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)=\dfrac{1}{n}\left(\dfrac{1}{2} m \omega^{2} x^{2}\right)\end{aligned}\) \(\Rightarrow x=\sqrt{\dfrac{n}{n+1}} A\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXI14:OSCILLATIONS
364094
The potential energy of a particle \(\left(U_{x}\right)\) executing SHM is given by
1 \(U_{x}=k_{1} x+k_{2} x^{2}+k_{3} x^{3}\)
2 \(U_{x}=\dfrac{k}{2}(x-a)^{2}\)
3 \(U_{x}=\) constant
4 \(U_{x}=A e^{-b x}\)
Explanation:
Potential energy of body in SHM at an instant, \(U=\dfrac{1}{2} k y^{2}\) where the displacement, \(y=(x-a)\)
PHXI14:OSCILLATIONS
364095
A particle of mass \(m\) moves in a one-dimensional potential energy \(U(x)=a x^{2}-b x^{4}\), where \(a\) and \(b\) are positive constants. The angular frequency of small oscillations about the minima of the potential energy is equal to
1 \(2 \sqrt{\dfrac{a}{m}}\)
2 \(\pi \sqrt{\dfrac{a}{2 b}}\)
3 \(\sqrt{\dfrac{a}{2 m}}\)
4 \(\sqrt{\dfrac{2 a}{m}}\)
Explanation:
\(U(x)=-a x^{2}+b x^{4}\) \(\therefore F=-\dfrac{d U}{d x}=-2 a x+4 b x^{3}\) At mean position \(F=0\) \(\begin{gathered}\therefore 2 a x=4 b x^{3} \\\text { or } x^{2}=\dfrac{2 a}{4 b}=\dfrac{a}{2 b} \Rightarrow x= \pm \sqrt{a / 2 b} \\\text { ' } \mathrm{U} \text { ' is minimum at } x=-\sqrt{\dfrac{a}{2 b}}\end{gathered}\) Now, \(F=m \dfrac{d^{2} x}{d t^{2}}\) \(\begin{aligned}& \therefore \quad m \dfrac{d^{2} x}{d t^{2}}=-2 a x+4 b x^{3} \\& \Rightarrow \dfrac{d^{2} x}{d t^{2}}=-\dfrac{2 a x}{m}\left(1-\dfrac{2 b}{a} x^{2}\right)\end{aligned}\) For small ' \(x\) ', \(\begin{aligned}& \dfrac{d^{2} x}{d t^{2}}=-\left(\dfrac{2 a}{m}\right) x \\& \therefore \quad \omega=\sqrt{2 a / m}\end{aligned}\)
PHXI14:OSCILLATIONS
364096
A body executes simple harmonic motion. The potential energy (PE), kinetic energy (KE) and total energy (TE) are measured as a function of displacement \(x\). Which of the following statement is true?
1 \(PE\) is maximum when \(x=0\)
2 \(TE\) is zero when \(x=0\)
3 \(KE\) is maximum when \(x\) is maximum
4 \(KE\) is maximum when \(x=0\)
Explanation:
\(P E=\dfrac{1}{2} m \omega^{2} x^{2}\) \(KE = \frac{1}{2}m{\omega ^2}\left( {{a^2} - {x^2}} \right)\) \(TE = PE + KE = \frac{1}{2}m{\omega ^2}{a^2}\) Since \(PE\) is maximum at \(x=a\) and \(KE\) is maximum at \(x=0\), therefore \(TE\) remains constant throughout the motion.
PHXI14:OSCILLATIONS
364097
The displacement of a particle executing SHM is given by \(y=5 \sin \left(4 t+\dfrac{\pi}{3}\right)\). If \(T\) is the time period and mass of the particle is \(2\;g\), the kinetic energy of the particle when \(t=\dfrac{T}{4}\) is given by
1 \(0.5\;J\)
2 \(0.4\;J\)
3 \(0.3\;J\)
4 \(3\;J\)
Explanation:
The displacement of particle, executing SHM, \(\begin{aligned}& y=5 \sin \left(4 t+\dfrac{\pi}{3}\right) \\& v=20 \cos \left(4 t+\dfrac{\pi}{3}\right)\end{aligned}\) Velocity at \(t=\left(\dfrac{T}{4}\right)\) \(\begin{aligned}& v=20 \cos \left(4 \times \dfrac{T}{4}+\dfrac{\pi}{3}\right) \\& \Rightarrow v=20 \cos \left(T+\dfrac{\pi}{3}\right)\end{aligned}\) Comparing the given equation with standard equation of SHM \(y=a \sin (\omega t+\phi)\), we get \(\omega=4\). \(\text { As } T=\dfrac{2 \pi}{\omega} \Rightarrow T=\dfrac{2 \pi}{4} \Rightarrow T=\dfrac{\pi}{2}\) Now, putting value of \(T\) in Eq. (2), we get \(\begin{aligned}v & =20 \cos \left(\dfrac{\pi}{2}+\dfrac{\pi}{3}\right)=-20 \sin \dfrac{\pi}{3} \\& =-20 \times \dfrac{\sqrt{3}}{2}=-10 \times \sqrt{3}\end{aligned}\) The kinetic energy of particle, \(KE = \frac{1}{2}m{v^2}\) \(\because m = 2g = 2 \times {10^{ - 3}}\;kg\) \( = \frac{1}{2} \times 2 \times {10^{ - 3}} \times {( - 10\sqrt 3 )^2}\) \( = {10^{ - 3}} \times 100 \times 3 = 3 \times {10^{ - 1}} \Rightarrow K.E = 0.3\,J.\)
PHXI14:OSCILLATIONS
364098
The kinetic energy of SHM is \(1 / n\) time its potential energy. If the amplitude of the SHM be \(A\), then what is the displacement of the particle?
1 \(\dfrac{A}{n}\)
2 \(\sqrt{\dfrac{n+1}{n}} A\)
3 \(\sqrt{\dfrac{n}{n+1}} A\)
4 \(n A\)
Explanation:
\(\begin{aligned}& K E=\dfrac{1}{n} P E \\& \dfrac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)=\dfrac{1}{n}\left(\dfrac{1}{2} m \omega^{2} x^{2}\right)\end{aligned}\) \(\Rightarrow x=\sqrt{\dfrac{n}{n+1}} A\)
364094
The potential energy of a particle \(\left(U_{x}\right)\) executing SHM is given by
1 \(U_{x}=k_{1} x+k_{2} x^{2}+k_{3} x^{3}\)
2 \(U_{x}=\dfrac{k}{2}(x-a)^{2}\)
3 \(U_{x}=\) constant
4 \(U_{x}=A e^{-b x}\)
Explanation:
Potential energy of body in SHM at an instant, \(U=\dfrac{1}{2} k y^{2}\) where the displacement, \(y=(x-a)\)
PHXI14:OSCILLATIONS
364095
A particle of mass \(m\) moves in a one-dimensional potential energy \(U(x)=a x^{2}-b x^{4}\), where \(a\) and \(b\) are positive constants. The angular frequency of small oscillations about the minima of the potential energy is equal to
1 \(2 \sqrt{\dfrac{a}{m}}\)
2 \(\pi \sqrt{\dfrac{a}{2 b}}\)
3 \(\sqrt{\dfrac{a}{2 m}}\)
4 \(\sqrt{\dfrac{2 a}{m}}\)
Explanation:
\(U(x)=-a x^{2}+b x^{4}\) \(\therefore F=-\dfrac{d U}{d x}=-2 a x+4 b x^{3}\) At mean position \(F=0\) \(\begin{gathered}\therefore 2 a x=4 b x^{3} \\\text { or } x^{2}=\dfrac{2 a}{4 b}=\dfrac{a}{2 b} \Rightarrow x= \pm \sqrt{a / 2 b} \\\text { ' } \mathrm{U} \text { ' is minimum at } x=-\sqrt{\dfrac{a}{2 b}}\end{gathered}\) Now, \(F=m \dfrac{d^{2} x}{d t^{2}}\) \(\begin{aligned}& \therefore \quad m \dfrac{d^{2} x}{d t^{2}}=-2 a x+4 b x^{3} \\& \Rightarrow \dfrac{d^{2} x}{d t^{2}}=-\dfrac{2 a x}{m}\left(1-\dfrac{2 b}{a} x^{2}\right)\end{aligned}\) For small ' \(x\) ', \(\begin{aligned}& \dfrac{d^{2} x}{d t^{2}}=-\left(\dfrac{2 a}{m}\right) x \\& \therefore \quad \omega=\sqrt{2 a / m}\end{aligned}\)
PHXI14:OSCILLATIONS
364096
A body executes simple harmonic motion. The potential energy (PE), kinetic energy (KE) and total energy (TE) are measured as a function of displacement \(x\). Which of the following statement is true?
1 \(PE\) is maximum when \(x=0\)
2 \(TE\) is zero when \(x=0\)
3 \(KE\) is maximum when \(x\) is maximum
4 \(KE\) is maximum when \(x=0\)
Explanation:
\(P E=\dfrac{1}{2} m \omega^{2} x^{2}\) \(KE = \frac{1}{2}m{\omega ^2}\left( {{a^2} - {x^2}} \right)\) \(TE = PE + KE = \frac{1}{2}m{\omega ^2}{a^2}\) Since \(PE\) is maximum at \(x=a\) and \(KE\) is maximum at \(x=0\), therefore \(TE\) remains constant throughout the motion.
PHXI14:OSCILLATIONS
364097
The displacement of a particle executing SHM is given by \(y=5 \sin \left(4 t+\dfrac{\pi}{3}\right)\). If \(T\) is the time period and mass of the particle is \(2\;g\), the kinetic energy of the particle when \(t=\dfrac{T}{4}\) is given by
1 \(0.5\;J\)
2 \(0.4\;J\)
3 \(0.3\;J\)
4 \(3\;J\)
Explanation:
The displacement of particle, executing SHM, \(\begin{aligned}& y=5 \sin \left(4 t+\dfrac{\pi}{3}\right) \\& v=20 \cos \left(4 t+\dfrac{\pi}{3}\right)\end{aligned}\) Velocity at \(t=\left(\dfrac{T}{4}\right)\) \(\begin{aligned}& v=20 \cos \left(4 \times \dfrac{T}{4}+\dfrac{\pi}{3}\right) \\& \Rightarrow v=20 \cos \left(T+\dfrac{\pi}{3}\right)\end{aligned}\) Comparing the given equation with standard equation of SHM \(y=a \sin (\omega t+\phi)\), we get \(\omega=4\). \(\text { As } T=\dfrac{2 \pi}{\omega} \Rightarrow T=\dfrac{2 \pi}{4} \Rightarrow T=\dfrac{\pi}{2}\) Now, putting value of \(T\) in Eq. (2), we get \(\begin{aligned}v & =20 \cos \left(\dfrac{\pi}{2}+\dfrac{\pi}{3}\right)=-20 \sin \dfrac{\pi}{3} \\& =-20 \times \dfrac{\sqrt{3}}{2}=-10 \times \sqrt{3}\end{aligned}\) The kinetic energy of particle, \(KE = \frac{1}{2}m{v^2}\) \(\because m = 2g = 2 \times {10^{ - 3}}\;kg\) \( = \frac{1}{2} \times 2 \times {10^{ - 3}} \times {( - 10\sqrt 3 )^2}\) \( = {10^{ - 3}} \times 100 \times 3 = 3 \times {10^{ - 1}} \Rightarrow K.E = 0.3\,J.\)
PHXI14:OSCILLATIONS
364098
The kinetic energy of SHM is \(1 / n\) time its potential energy. If the amplitude of the SHM be \(A\), then what is the displacement of the particle?
1 \(\dfrac{A}{n}\)
2 \(\sqrt{\dfrac{n+1}{n}} A\)
3 \(\sqrt{\dfrac{n}{n+1}} A\)
4 \(n A\)
Explanation:
\(\begin{aligned}& K E=\dfrac{1}{n} P E \\& \dfrac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)=\dfrac{1}{n}\left(\dfrac{1}{2} m \omega^{2} x^{2}\right)\end{aligned}\) \(\Rightarrow x=\sqrt{\dfrac{n}{n+1}} A\)
364094
The potential energy of a particle \(\left(U_{x}\right)\) executing SHM is given by
1 \(U_{x}=k_{1} x+k_{2} x^{2}+k_{3} x^{3}\)
2 \(U_{x}=\dfrac{k}{2}(x-a)^{2}\)
3 \(U_{x}=\) constant
4 \(U_{x}=A e^{-b x}\)
Explanation:
Potential energy of body in SHM at an instant, \(U=\dfrac{1}{2} k y^{2}\) where the displacement, \(y=(x-a)\)
PHXI14:OSCILLATIONS
364095
A particle of mass \(m\) moves in a one-dimensional potential energy \(U(x)=a x^{2}-b x^{4}\), where \(a\) and \(b\) are positive constants. The angular frequency of small oscillations about the minima of the potential energy is equal to
1 \(2 \sqrt{\dfrac{a}{m}}\)
2 \(\pi \sqrt{\dfrac{a}{2 b}}\)
3 \(\sqrt{\dfrac{a}{2 m}}\)
4 \(\sqrt{\dfrac{2 a}{m}}\)
Explanation:
\(U(x)=-a x^{2}+b x^{4}\) \(\therefore F=-\dfrac{d U}{d x}=-2 a x+4 b x^{3}\) At mean position \(F=0\) \(\begin{gathered}\therefore 2 a x=4 b x^{3} \\\text { or } x^{2}=\dfrac{2 a}{4 b}=\dfrac{a}{2 b} \Rightarrow x= \pm \sqrt{a / 2 b} \\\text { ' } \mathrm{U} \text { ' is minimum at } x=-\sqrt{\dfrac{a}{2 b}}\end{gathered}\) Now, \(F=m \dfrac{d^{2} x}{d t^{2}}\) \(\begin{aligned}& \therefore \quad m \dfrac{d^{2} x}{d t^{2}}=-2 a x+4 b x^{3} \\& \Rightarrow \dfrac{d^{2} x}{d t^{2}}=-\dfrac{2 a x}{m}\left(1-\dfrac{2 b}{a} x^{2}\right)\end{aligned}\) For small ' \(x\) ', \(\begin{aligned}& \dfrac{d^{2} x}{d t^{2}}=-\left(\dfrac{2 a}{m}\right) x \\& \therefore \quad \omega=\sqrt{2 a / m}\end{aligned}\)
PHXI14:OSCILLATIONS
364096
A body executes simple harmonic motion. The potential energy (PE), kinetic energy (KE) and total energy (TE) are measured as a function of displacement \(x\). Which of the following statement is true?
1 \(PE\) is maximum when \(x=0\)
2 \(TE\) is zero when \(x=0\)
3 \(KE\) is maximum when \(x\) is maximum
4 \(KE\) is maximum when \(x=0\)
Explanation:
\(P E=\dfrac{1}{2} m \omega^{2} x^{2}\) \(KE = \frac{1}{2}m{\omega ^2}\left( {{a^2} - {x^2}} \right)\) \(TE = PE + KE = \frac{1}{2}m{\omega ^2}{a^2}\) Since \(PE\) is maximum at \(x=a\) and \(KE\) is maximum at \(x=0\), therefore \(TE\) remains constant throughout the motion.
PHXI14:OSCILLATIONS
364097
The displacement of a particle executing SHM is given by \(y=5 \sin \left(4 t+\dfrac{\pi}{3}\right)\). If \(T\) is the time period and mass of the particle is \(2\;g\), the kinetic energy of the particle when \(t=\dfrac{T}{4}\) is given by
1 \(0.5\;J\)
2 \(0.4\;J\)
3 \(0.3\;J\)
4 \(3\;J\)
Explanation:
The displacement of particle, executing SHM, \(\begin{aligned}& y=5 \sin \left(4 t+\dfrac{\pi}{3}\right) \\& v=20 \cos \left(4 t+\dfrac{\pi}{3}\right)\end{aligned}\) Velocity at \(t=\left(\dfrac{T}{4}\right)\) \(\begin{aligned}& v=20 \cos \left(4 \times \dfrac{T}{4}+\dfrac{\pi}{3}\right) \\& \Rightarrow v=20 \cos \left(T+\dfrac{\pi}{3}\right)\end{aligned}\) Comparing the given equation with standard equation of SHM \(y=a \sin (\omega t+\phi)\), we get \(\omega=4\). \(\text { As } T=\dfrac{2 \pi}{\omega} \Rightarrow T=\dfrac{2 \pi}{4} \Rightarrow T=\dfrac{\pi}{2}\) Now, putting value of \(T\) in Eq. (2), we get \(\begin{aligned}v & =20 \cos \left(\dfrac{\pi}{2}+\dfrac{\pi}{3}\right)=-20 \sin \dfrac{\pi}{3} \\& =-20 \times \dfrac{\sqrt{3}}{2}=-10 \times \sqrt{3}\end{aligned}\) The kinetic energy of particle, \(KE = \frac{1}{2}m{v^2}\) \(\because m = 2g = 2 \times {10^{ - 3}}\;kg\) \( = \frac{1}{2} \times 2 \times {10^{ - 3}} \times {( - 10\sqrt 3 )^2}\) \( = {10^{ - 3}} \times 100 \times 3 = 3 \times {10^{ - 1}} \Rightarrow K.E = 0.3\,J.\)
PHXI14:OSCILLATIONS
364098
The kinetic energy of SHM is \(1 / n\) time its potential energy. If the amplitude of the SHM be \(A\), then what is the displacement of the particle?
1 \(\dfrac{A}{n}\)
2 \(\sqrt{\dfrac{n+1}{n}} A\)
3 \(\sqrt{\dfrac{n}{n+1}} A\)
4 \(n A\)
Explanation:
\(\begin{aligned}& K E=\dfrac{1}{n} P E \\& \dfrac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)=\dfrac{1}{n}\left(\dfrac{1}{2} m \omega^{2} x^{2}\right)\end{aligned}\) \(\Rightarrow x=\sqrt{\dfrac{n}{n+1}} A\)