364245
The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of \(1.0\;m\). If the piston moves with simple harmonic motion with an angular frequency of \(200\,rad/\min \), what is its maximum speed?
1 \(200\;m/\min \)
2 \(100\;m/\min \)
3 \(50\;m/\min \)
4 \(300\;m/\min \)
Explanation:
Given, angular frequency of the piston, \(\omega = 200\,rad/\min \) Stroke length \( = 1\;m\) \(\therefore\) Amplitude of SHM, \(A = \frac{{{\rm{ }}Stroke{\rm{ }}length{\rm{ }}}}{2} = \frac{1}{2} = 0.5\;m\) Now, \({v_{\max }} = \omega A = 200 \times 0.5 = 100\;m/\min \)
NCERT Exemplar
PHXI14:OSCILLATIONS
364246
If a particle executes SHM with angular velocity \(3.5\,rad{\rm{/}}s\) and maximum acceleration \(7.5\;m{\rm{/}}{s^2}\), then the amplitude of oscillation will be
1 \(0.8\;m\)
2 \(0.69\;m\)
3 \(0.41\;m\)
4 \(0.61\;m\)
Explanation:
Maximum acceleration for particle executing SHM. \(a_{\max }=\omega^{2} A\) \(A=\dfrac{a_{\max }}{\omega^{2}}\) \(A = \frac{{7.5}}{{{{(3.5)}^2}}} \Rightarrow 0.61\;m\)
PHXI14:OSCILLATIONS
364247
The variation of velocity of a particle executing SHM with time is shown in fig. The velocity of the particle when a phase change of \(\pi / 6\) takes place from the instant it is at one of the extreme positions will be
1 \(2.5\;m/s\)
2 \(3.53\;m/s\)
3 \(4.330\;m/s\)
4 None of these
Explanation:
From the graph \(T = 4\;s\) and \({v_{\max {\rm{ }}}} = 5\;m/s;\;\;\;{\mkern 1mu} {\kern 1pt} \omega A = 5\;m/s\) \(\left(\dfrac{2 \pi}{T}\right) A=5 \Rightarrow A=\dfrac{5 T}{2 \pi}=\dfrac{5 \times 4}{2 \pi}=\dfrac{10}{\pi} m\) Also, \(\omega = \frac{{2\pi }}{4} = \frac{\pi }{2}rad/s\) The equation of velocity can be written as \(v=5 \sin \left(\dfrac{\pi}{2} t\right) m / s\) The phase \(\phi=\dfrac{\pi}{2} t\) Phase change is \(\Delta \phi = \frac{\pi }{2}\Delta t = \frac{\pi }{6} \Rightarrow \Delta t = {t_f} - {t_i} = t = \frac{1}{3}s\) \(v = 5\sin \left( {\frac{\pi }{2}\left( {\frac{1}{3}} \right)} \right) = 2.5\;m/s\)
PHXI14:OSCILLATIONS
364248
The amplitude of a particle executing S.H.M with frequency of \(60\;Hz\) is \(0.01\;m\). the maximum value of the acceleration of the particle is
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PHXI14:OSCILLATIONS
364245
The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of \(1.0\;m\). If the piston moves with simple harmonic motion with an angular frequency of \(200\,rad/\min \), what is its maximum speed?
1 \(200\;m/\min \)
2 \(100\;m/\min \)
3 \(50\;m/\min \)
4 \(300\;m/\min \)
Explanation:
Given, angular frequency of the piston, \(\omega = 200\,rad/\min \) Stroke length \( = 1\;m\) \(\therefore\) Amplitude of SHM, \(A = \frac{{{\rm{ }}Stroke{\rm{ }}length{\rm{ }}}}{2} = \frac{1}{2} = 0.5\;m\) Now, \({v_{\max }} = \omega A = 200 \times 0.5 = 100\;m/\min \)
NCERT Exemplar
PHXI14:OSCILLATIONS
364246
If a particle executes SHM with angular velocity \(3.5\,rad{\rm{/}}s\) and maximum acceleration \(7.5\;m{\rm{/}}{s^2}\), then the amplitude of oscillation will be
1 \(0.8\;m\)
2 \(0.69\;m\)
3 \(0.41\;m\)
4 \(0.61\;m\)
Explanation:
Maximum acceleration for particle executing SHM. \(a_{\max }=\omega^{2} A\) \(A=\dfrac{a_{\max }}{\omega^{2}}\) \(A = \frac{{7.5}}{{{{(3.5)}^2}}} \Rightarrow 0.61\;m\)
PHXI14:OSCILLATIONS
364247
The variation of velocity of a particle executing SHM with time is shown in fig. The velocity of the particle when a phase change of \(\pi / 6\) takes place from the instant it is at one of the extreme positions will be
1 \(2.5\;m/s\)
2 \(3.53\;m/s\)
3 \(4.330\;m/s\)
4 None of these
Explanation:
From the graph \(T = 4\;s\) and \({v_{\max {\rm{ }}}} = 5\;m/s;\;\;\;{\mkern 1mu} {\kern 1pt} \omega A = 5\;m/s\) \(\left(\dfrac{2 \pi}{T}\right) A=5 \Rightarrow A=\dfrac{5 T}{2 \pi}=\dfrac{5 \times 4}{2 \pi}=\dfrac{10}{\pi} m\) Also, \(\omega = \frac{{2\pi }}{4} = \frac{\pi }{2}rad/s\) The equation of velocity can be written as \(v=5 \sin \left(\dfrac{\pi}{2} t\right) m / s\) The phase \(\phi=\dfrac{\pi}{2} t\) Phase change is \(\Delta \phi = \frac{\pi }{2}\Delta t = \frac{\pi }{6} \Rightarrow \Delta t = {t_f} - {t_i} = t = \frac{1}{3}s\) \(v = 5\sin \left( {\frac{\pi }{2}\left( {\frac{1}{3}} \right)} \right) = 2.5\;m/s\)
PHXI14:OSCILLATIONS
364248
The amplitude of a particle executing S.H.M with frequency of \(60\;Hz\) is \(0.01\;m\). the maximum value of the acceleration of the particle is
364245
The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of \(1.0\;m\). If the piston moves with simple harmonic motion with an angular frequency of \(200\,rad/\min \), what is its maximum speed?
1 \(200\;m/\min \)
2 \(100\;m/\min \)
3 \(50\;m/\min \)
4 \(300\;m/\min \)
Explanation:
Given, angular frequency of the piston, \(\omega = 200\,rad/\min \) Stroke length \( = 1\;m\) \(\therefore\) Amplitude of SHM, \(A = \frac{{{\rm{ }}Stroke{\rm{ }}length{\rm{ }}}}{2} = \frac{1}{2} = 0.5\;m\) Now, \({v_{\max }} = \omega A = 200 \times 0.5 = 100\;m/\min \)
NCERT Exemplar
PHXI14:OSCILLATIONS
364246
If a particle executes SHM with angular velocity \(3.5\,rad{\rm{/}}s\) and maximum acceleration \(7.5\;m{\rm{/}}{s^2}\), then the amplitude of oscillation will be
1 \(0.8\;m\)
2 \(0.69\;m\)
3 \(0.41\;m\)
4 \(0.61\;m\)
Explanation:
Maximum acceleration for particle executing SHM. \(a_{\max }=\omega^{2} A\) \(A=\dfrac{a_{\max }}{\omega^{2}}\) \(A = \frac{{7.5}}{{{{(3.5)}^2}}} \Rightarrow 0.61\;m\)
PHXI14:OSCILLATIONS
364247
The variation of velocity of a particle executing SHM with time is shown in fig. The velocity of the particle when a phase change of \(\pi / 6\) takes place from the instant it is at one of the extreme positions will be
1 \(2.5\;m/s\)
2 \(3.53\;m/s\)
3 \(4.330\;m/s\)
4 None of these
Explanation:
From the graph \(T = 4\;s\) and \({v_{\max {\rm{ }}}} = 5\;m/s;\;\;\;{\mkern 1mu} {\kern 1pt} \omega A = 5\;m/s\) \(\left(\dfrac{2 \pi}{T}\right) A=5 \Rightarrow A=\dfrac{5 T}{2 \pi}=\dfrac{5 \times 4}{2 \pi}=\dfrac{10}{\pi} m\) Also, \(\omega = \frac{{2\pi }}{4} = \frac{\pi }{2}rad/s\) The equation of velocity can be written as \(v=5 \sin \left(\dfrac{\pi}{2} t\right) m / s\) The phase \(\phi=\dfrac{\pi}{2} t\) Phase change is \(\Delta \phi = \frac{\pi }{2}\Delta t = \frac{\pi }{6} \Rightarrow \Delta t = {t_f} - {t_i} = t = \frac{1}{3}s\) \(v = 5\sin \left( {\frac{\pi }{2}\left( {\frac{1}{3}} \right)} \right) = 2.5\;m/s\)
PHXI14:OSCILLATIONS
364248
The amplitude of a particle executing S.H.M with frequency of \(60\;Hz\) is \(0.01\;m\). the maximum value of the acceleration of the particle is
364245
The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of \(1.0\;m\). If the piston moves with simple harmonic motion with an angular frequency of \(200\,rad/\min \), what is its maximum speed?
1 \(200\;m/\min \)
2 \(100\;m/\min \)
3 \(50\;m/\min \)
4 \(300\;m/\min \)
Explanation:
Given, angular frequency of the piston, \(\omega = 200\,rad/\min \) Stroke length \( = 1\;m\) \(\therefore\) Amplitude of SHM, \(A = \frac{{{\rm{ }}Stroke{\rm{ }}length{\rm{ }}}}{2} = \frac{1}{2} = 0.5\;m\) Now, \({v_{\max }} = \omega A = 200 \times 0.5 = 100\;m/\min \)
NCERT Exemplar
PHXI14:OSCILLATIONS
364246
If a particle executes SHM with angular velocity \(3.5\,rad{\rm{/}}s\) and maximum acceleration \(7.5\;m{\rm{/}}{s^2}\), then the amplitude of oscillation will be
1 \(0.8\;m\)
2 \(0.69\;m\)
3 \(0.41\;m\)
4 \(0.61\;m\)
Explanation:
Maximum acceleration for particle executing SHM. \(a_{\max }=\omega^{2} A\) \(A=\dfrac{a_{\max }}{\omega^{2}}\) \(A = \frac{{7.5}}{{{{(3.5)}^2}}} \Rightarrow 0.61\;m\)
PHXI14:OSCILLATIONS
364247
The variation of velocity of a particle executing SHM with time is shown in fig. The velocity of the particle when a phase change of \(\pi / 6\) takes place from the instant it is at one of the extreme positions will be
1 \(2.5\;m/s\)
2 \(3.53\;m/s\)
3 \(4.330\;m/s\)
4 None of these
Explanation:
From the graph \(T = 4\;s\) and \({v_{\max {\rm{ }}}} = 5\;m/s;\;\;\;{\mkern 1mu} {\kern 1pt} \omega A = 5\;m/s\) \(\left(\dfrac{2 \pi}{T}\right) A=5 \Rightarrow A=\dfrac{5 T}{2 \pi}=\dfrac{5 \times 4}{2 \pi}=\dfrac{10}{\pi} m\) Also, \(\omega = \frac{{2\pi }}{4} = \frac{\pi }{2}rad/s\) The equation of velocity can be written as \(v=5 \sin \left(\dfrac{\pi}{2} t\right) m / s\) The phase \(\phi=\dfrac{\pi}{2} t\) Phase change is \(\Delta \phi = \frac{\pi }{2}\Delta t = \frac{\pi }{6} \Rightarrow \Delta t = {t_f} - {t_i} = t = \frac{1}{3}s\) \(v = 5\sin \left( {\frac{\pi }{2}\left( {\frac{1}{3}} \right)} \right) = 2.5\;m/s\)
PHXI14:OSCILLATIONS
364248
The amplitude of a particle executing S.H.M with frequency of \(60\;Hz\) is \(0.01\;m\). the maximum value of the acceleration of the particle is
