364189
A particle executes SHM with amplitude \(0.2\;m\) and time period \(24\;s\). The time required for it to move from the mean position to a point \(0.1\;m\) from the mean position is
1 \(2\;s\)
2 \(3\;s\)
3 \(8\;s\)
4 \(12\;s\)
Explanation:
Given that \(A = 0.2\;m,\;T = 24\;s\) Displacement \(x\) of a particle from its mean position is given by \(x = {\rm{ }}A\sin {\rm{ }}\omega t\) \({\rm{Here}},{\rm{ }}x = 0.1\;m\) \(\therefore 0.1 = 0.2\sin \omega t\) \(\frac{1}{2} = \sin \omega t\) \( \Rightarrow \omega t = \frac{\pi }{6}t = \frac{\pi }{{6\omega }} = \frac{\pi }{6}\left( {\frac{T}{{2\pi }}} \right)\;\;\;{\mkern 1mu} {\kern 1pt} \left( {\because \omega = \frac{{2\pi }}{T}} \right)\) \( = \frac{{\pi \times 24}}{{6 \times 2\pi }} = 2\;s\)
KCET - 2012
PHXI14:OSCILLATIONS
364190
A particle excuting a simple harmonic motion has a period of \(6{\rm{ }}\sec \). The time taken by the particle to move from the mean position to half the amplitude, starting from the mean position is
1 \(\frac{3}{2}\,\sec \)
2 \(\frac{1}{2}\,\sec \)
3 \(\frac{3}{4}\,\sec \)
4 \(\frac{1}{4}\,\sec \)
Explanation:
Since the motion is started from the mean position, of the particle is given by \(x = A\sin \omega t\) \(At{\rm{ }}x = \frac{A}{2}\) \(\therefore \frac{A}{2} = A\sin \omega t \Rightarrow \frac{1}{2} = \sin \omega t \Rightarrow \frac{\pi }{6} = \omega t\) \(t = \frac{\pi }{{6\omega }} = \frac{\pi }{{6\left( {\frac{{2\pi }}{T}} \right)}}\;\;\;{\mkern 1mu} {\kern 1pt} \left( {\because \omega = \frac{{2\pi }}{T}} \right)\) \( = \frac{T}{{12}} = \frac{6}{{12}} = \frac{1}{2}s\)
KCET - 2011
PHXI14:OSCILLATIONS
364191
A particle executes the simple harmonic motion with an amplitude 'A'. The distance travelled by it in one periodic time is
1 \(\dfrac{1}{2}\)
2 \(A\)
3 \(2\;A\)
4 \(4\;A\)
Explanation:
In simple harmonic motion total distance travelled is \(4\;A\). This is because the particle covers \(A\) in \(\dfrac{T}{4}\) time period so in complete \(T\) time it covers 4 times the amplitude distance i.e., \(4A\)
PHXI14:OSCILLATIONS
364192
A particle executes simple harmonic oscillation with an amplitude \(a\). The period of oscillation is \(T\). The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is
1 \(\dfrac{T}{8}\)
2 \(\dfrac{T}{4}\)
3 \(\dfrac{T}{2}\)
4 \(\dfrac{T}{12}\)
Explanation:
Let the displacement equation of particle executing SHM is \(y=a \sin \omega t\) As particle travels half of the amplitude from the equilibrium position, so \(y=\dfrac{a}{2}\) Therefore, \(\dfrac{a}{2}=a \sin \omega t\) or \(\sin \omega t=\dfrac{1}{2}=\sin \dfrac{\pi}{6}\) \(\Rightarrow t=\dfrac{T}{12}\)
364189
A particle executes SHM with amplitude \(0.2\;m\) and time period \(24\;s\). The time required for it to move from the mean position to a point \(0.1\;m\) from the mean position is
1 \(2\;s\)
2 \(3\;s\)
3 \(8\;s\)
4 \(12\;s\)
Explanation:
Given that \(A = 0.2\;m,\;T = 24\;s\) Displacement \(x\) of a particle from its mean position is given by \(x = {\rm{ }}A\sin {\rm{ }}\omega t\) \({\rm{Here}},{\rm{ }}x = 0.1\;m\) \(\therefore 0.1 = 0.2\sin \omega t\) \(\frac{1}{2} = \sin \omega t\) \( \Rightarrow \omega t = \frac{\pi }{6}t = \frac{\pi }{{6\omega }} = \frac{\pi }{6}\left( {\frac{T}{{2\pi }}} \right)\;\;\;{\mkern 1mu} {\kern 1pt} \left( {\because \omega = \frac{{2\pi }}{T}} \right)\) \( = \frac{{\pi \times 24}}{{6 \times 2\pi }} = 2\;s\)
KCET - 2012
PHXI14:OSCILLATIONS
364190
A particle excuting a simple harmonic motion has a period of \(6{\rm{ }}\sec \). The time taken by the particle to move from the mean position to half the amplitude, starting from the mean position is
1 \(\frac{3}{2}\,\sec \)
2 \(\frac{1}{2}\,\sec \)
3 \(\frac{3}{4}\,\sec \)
4 \(\frac{1}{4}\,\sec \)
Explanation:
Since the motion is started from the mean position, of the particle is given by \(x = A\sin \omega t\) \(At{\rm{ }}x = \frac{A}{2}\) \(\therefore \frac{A}{2} = A\sin \omega t \Rightarrow \frac{1}{2} = \sin \omega t \Rightarrow \frac{\pi }{6} = \omega t\) \(t = \frac{\pi }{{6\omega }} = \frac{\pi }{{6\left( {\frac{{2\pi }}{T}} \right)}}\;\;\;{\mkern 1mu} {\kern 1pt} \left( {\because \omega = \frac{{2\pi }}{T}} \right)\) \( = \frac{T}{{12}} = \frac{6}{{12}} = \frac{1}{2}s\)
KCET - 2011
PHXI14:OSCILLATIONS
364191
A particle executes the simple harmonic motion with an amplitude 'A'. The distance travelled by it in one periodic time is
1 \(\dfrac{1}{2}\)
2 \(A\)
3 \(2\;A\)
4 \(4\;A\)
Explanation:
In simple harmonic motion total distance travelled is \(4\;A\). This is because the particle covers \(A\) in \(\dfrac{T}{4}\) time period so in complete \(T\) time it covers 4 times the amplitude distance i.e., \(4A\)
PHXI14:OSCILLATIONS
364192
A particle executes simple harmonic oscillation with an amplitude \(a\). The period of oscillation is \(T\). The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is
1 \(\dfrac{T}{8}\)
2 \(\dfrac{T}{4}\)
3 \(\dfrac{T}{2}\)
4 \(\dfrac{T}{12}\)
Explanation:
Let the displacement equation of particle executing SHM is \(y=a \sin \omega t\) As particle travels half of the amplitude from the equilibrium position, so \(y=\dfrac{a}{2}\) Therefore, \(\dfrac{a}{2}=a \sin \omega t\) or \(\sin \omega t=\dfrac{1}{2}=\sin \dfrac{\pi}{6}\) \(\Rightarrow t=\dfrac{T}{12}\)
364189
A particle executes SHM with amplitude \(0.2\;m\) and time period \(24\;s\). The time required for it to move from the mean position to a point \(0.1\;m\) from the mean position is
1 \(2\;s\)
2 \(3\;s\)
3 \(8\;s\)
4 \(12\;s\)
Explanation:
Given that \(A = 0.2\;m,\;T = 24\;s\) Displacement \(x\) of a particle from its mean position is given by \(x = {\rm{ }}A\sin {\rm{ }}\omega t\) \({\rm{Here}},{\rm{ }}x = 0.1\;m\) \(\therefore 0.1 = 0.2\sin \omega t\) \(\frac{1}{2} = \sin \omega t\) \( \Rightarrow \omega t = \frac{\pi }{6}t = \frac{\pi }{{6\omega }} = \frac{\pi }{6}\left( {\frac{T}{{2\pi }}} \right)\;\;\;{\mkern 1mu} {\kern 1pt} \left( {\because \omega = \frac{{2\pi }}{T}} \right)\) \( = \frac{{\pi \times 24}}{{6 \times 2\pi }} = 2\;s\)
KCET - 2012
PHXI14:OSCILLATIONS
364190
A particle excuting a simple harmonic motion has a period of \(6{\rm{ }}\sec \). The time taken by the particle to move from the mean position to half the amplitude, starting from the mean position is
1 \(\frac{3}{2}\,\sec \)
2 \(\frac{1}{2}\,\sec \)
3 \(\frac{3}{4}\,\sec \)
4 \(\frac{1}{4}\,\sec \)
Explanation:
Since the motion is started from the mean position, of the particle is given by \(x = A\sin \omega t\) \(At{\rm{ }}x = \frac{A}{2}\) \(\therefore \frac{A}{2} = A\sin \omega t \Rightarrow \frac{1}{2} = \sin \omega t \Rightarrow \frac{\pi }{6} = \omega t\) \(t = \frac{\pi }{{6\omega }} = \frac{\pi }{{6\left( {\frac{{2\pi }}{T}} \right)}}\;\;\;{\mkern 1mu} {\kern 1pt} \left( {\because \omega = \frac{{2\pi }}{T}} \right)\) \( = \frac{T}{{12}} = \frac{6}{{12}} = \frac{1}{2}s\)
KCET - 2011
PHXI14:OSCILLATIONS
364191
A particle executes the simple harmonic motion with an amplitude 'A'. The distance travelled by it in one periodic time is
1 \(\dfrac{1}{2}\)
2 \(A\)
3 \(2\;A\)
4 \(4\;A\)
Explanation:
In simple harmonic motion total distance travelled is \(4\;A\). This is because the particle covers \(A\) in \(\dfrac{T}{4}\) time period so in complete \(T\) time it covers 4 times the amplitude distance i.e., \(4A\)
PHXI14:OSCILLATIONS
364192
A particle executes simple harmonic oscillation with an amplitude \(a\). The period of oscillation is \(T\). The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is
1 \(\dfrac{T}{8}\)
2 \(\dfrac{T}{4}\)
3 \(\dfrac{T}{2}\)
4 \(\dfrac{T}{12}\)
Explanation:
Let the displacement equation of particle executing SHM is \(y=a \sin \omega t\) As particle travels half of the amplitude from the equilibrium position, so \(y=\dfrac{a}{2}\) Therefore, \(\dfrac{a}{2}=a \sin \omega t\) or \(\sin \omega t=\dfrac{1}{2}=\sin \dfrac{\pi}{6}\) \(\Rightarrow t=\dfrac{T}{12}\)
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PHXI14:OSCILLATIONS
364189
A particle executes SHM with amplitude \(0.2\;m\) and time period \(24\;s\). The time required for it to move from the mean position to a point \(0.1\;m\) from the mean position is
1 \(2\;s\)
2 \(3\;s\)
3 \(8\;s\)
4 \(12\;s\)
Explanation:
Given that \(A = 0.2\;m,\;T = 24\;s\) Displacement \(x\) of a particle from its mean position is given by \(x = {\rm{ }}A\sin {\rm{ }}\omega t\) \({\rm{Here}},{\rm{ }}x = 0.1\;m\) \(\therefore 0.1 = 0.2\sin \omega t\) \(\frac{1}{2} = \sin \omega t\) \( \Rightarrow \omega t = \frac{\pi }{6}t = \frac{\pi }{{6\omega }} = \frac{\pi }{6}\left( {\frac{T}{{2\pi }}} \right)\;\;\;{\mkern 1mu} {\kern 1pt} \left( {\because \omega = \frac{{2\pi }}{T}} \right)\) \( = \frac{{\pi \times 24}}{{6 \times 2\pi }} = 2\;s\)
KCET - 2012
PHXI14:OSCILLATIONS
364190
A particle excuting a simple harmonic motion has a period of \(6{\rm{ }}\sec \). The time taken by the particle to move from the mean position to half the amplitude, starting from the mean position is
1 \(\frac{3}{2}\,\sec \)
2 \(\frac{1}{2}\,\sec \)
3 \(\frac{3}{4}\,\sec \)
4 \(\frac{1}{4}\,\sec \)
Explanation:
Since the motion is started from the mean position, of the particle is given by \(x = A\sin \omega t\) \(At{\rm{ }}x = \frac{A}{2}\) \(\therefore \frac{A}{2} = A\sin \omega t \Rightarrow \frac{1}{2} = \sin \omega t \Rightarrow \frac{\pi }{6} = \omega t\) \(t = \frac{\pi }{{6\omega }} = \frac{\pi }{{6\left( {\frac{{2\pi }}{T}} \right)}}\;\;\;{\mkern 1mu} {\kern 1pt} \left( {\because \omega = \frac{{2\pi }}{T}} \right)\) \( = \frac{T}{{12}} = \frac{6}{{12}} = \frac{1}{2}s\)
KCET - 2011
PHXI14:OSCILLATIONS
364191
A particle executes the simple harmonic motion with an amplitude 'A'. The distance travelled by it in one periodic time is
1 \(\dfrac{1}{2}\)
2 \(A\)
3 \(2\;A\)
4 \(4\;A\)
Explanation:
In simple harmonic motion total distance travelled is \(4\;A\). This is because the particle covers \(A\) in \(\dfrac{T}{4}\) time period so in complete \(T\) time it covers 4 times the amplitude distance i.e., \(4A\)
PHXI14:OSCILLATIONS
364192
A particle executes simple harmonic oscillation with an amplitude \(a\). The period of oscillation is \(T\). The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is
1 \(\dfrac{T}{8}\)
2 \(\dfrac{T}{4}\)
3 \(\dfrac{T}{2}\)
4 \(\dfrac{T}{12}\)
Explanation:
Let the displacement equation of particle executing SHM is \(y=a \sin \omega t\) As particle travels half of the amplitude from the equilibrium position, so \(y=\dfrac{a}{2}\) Therefore, \(\dfrac{a}{2}=a \sin \omega t\) or \(\sin \omega t=\dfrac{1}{2}=\sin \dfrac{\pi}{6}\) \(\Rightarrow t=\dfrac{T}{12}\)
