364176
Two simple harmonic motions are represented by \({y_1} = 5\left[ {\sin 2\pi \,t + \sqrt 3 \cos 2\pi \,t} \right]\) and \({y_2} = 5\sin \left( {2\pi \,t + \frac{\pi }{4}} \right).\) The ratio of their amplitudes is
364177
Statement A : If the amplitude of SHM of a spring mass system is increased, then time period of SHM will remain constant. Statement B : If amplitude is increased, body will have to travel more distance to complete one oscillation.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both statements are correct.
4 Both Statements are incorrect.
Explanation:
Time period of oscillation of spring mass system is \(T=2 \pi \sqrt{\dfrac{m}{k}}\) which is independent of the amplitude. Thus, if the amplitude of the system is increased, then \(T\) will remain same. So option (1) is correct.
PHXI14:OSCILLATIONS
364178
A particle performs SHM with amplitude \(25\;cm\) and period \(3\;s\). The minimum time required for it to move between two points \(12.5\;cm\) on either side of the mean position is
1 \(0.6\;s\)
2 \(0.5\;s\)
3 \(0.4\;s\)
4 \(0.2\;s\)
Explanation:
Here, amplitude of particle, \(A = 25\;cm\) and time period, \(T = 3\;s\) If the particle at \(t = 0\) is at mean position, its displacement equation will be \(x=A \sin \omega t\) i.e., \(x=25 \sin \dfrac{2 \pi t}{3}\left[\because \omega=2 \pi v=\dfrac{2 \pi}{T}=\dfrac{2 \pi}{3}\right]\) If it takes time \(t_{1}\) to move a distance \(x = 12.5\;cm\) to one side of its mean position, then \(12.5=25 \sin \dfrac{2 \pi t_{1}}{3}\) or \(\dfrac{1}{2}=\sin \dfrac{2 \pi t_{1}}{3}\) or \(\sin \dfrac{\pi}{6}=\sin \dfrac{2 \pi t_{1}}{3}\) \(\therefore \dfrac{\pi}{6}=\dfrac{2 \pi t_{1}}{3} \Rightarrow t_{1}=\dfrac{1}{4} s\) The same will be the time to move \(12.5\;cm\) to the other side of its mean position, therefore, total time \(t = {t_1} + {t_2} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} = 0.5\;s\)
PHXI14:OSCILLATIONS
364179
If \({x=5 \sin \left(\pi t+\dfrac{\pi}{3}\right) m}\) represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are
1 \(5\;cm,2\;s\)
2 \(5\,m,2\,s\)
3 \(5\;cm,1\;s\)
4 \(5\,m,1\,s\)
Explanation:
\({x=5 \sin \left(\pi t+\dfrac{\pi}{3}\right) m}\) comparing the given equation with standard one \({x=A \sin (\omega t+\phi)}\) Amplitude \({=5 m}\) \({\omega=\pi=\dfrac{2 \pi}{T}}\) \({T=\dfrac{2 \pi}{\pi}=2 s}\)
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PHXI14:OSCILLATIONS
364176
Two simple harmonic motions are represented by \({y_1} = 5\left[ {\sin 2\pi \,t + \sqrt 3 \cos 2\pi \,t} \right]\) and \({y_2} = 5\sin \left( {2\pi \,t + \frac{\pi }{4}} \right).\) The ratio of their amplitudes is
364177
Statement A : If the amplitude of SHM of a spring mass system is increased, then time period of SHM will remain constant. Statement B : If amplitude is increased, body will have to travel more distance to complete one oscillation.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both statements are correct.
4 Both Statements are incorrect.
Explanation:
Time period of oscillation of spring mass system is \(T=2 \pi \sqrt{\dfrac{m}{k}}\) which is independent of the amplitude. Thus, if the amplitude of the system is increased, then \(T\) will remain same. So option (1) is correct.
PHXI14:OSCILLATIONS
364178
A particle performs SHM with amplitude \(25\;cm\) and period \(3\;s\). The minimum time required for it to move between two points \(12.5\;cm\) on either side of the mean position is
1 \(0.6\;s\)
2 \(0.5\;s\)
3 \(0.4\;s\)
4 \(0.2\;s\)
Explanation:
Here, amplitude of particle, \(A = 25\;cm\) and time period, \(T = 3\;s\) If the particle at \(t = 0\) is at mean position, its displacement equation will be \(x=A \sin \omega t\) i.e., \(x=25 \sin \dfrac{2 \pi t}{3}\left[\because \omega=2 \pi v=\dfrac{2 \pi}{T}=\dfrac{2 \pi}{3}\right]\) If it takes time \(t_{1}\) to move a distance \(x = 12.5\;cm\) to one side of its mean position, then \(12.5=25 \sin \dfrac{2 \pi t_{1}}{3}\) or \(\dfrac{1}{2}=\sin \dfrac{2 \pi t_{1}}{3}\) or \(\sin \dfrac{\pi}{6}=\sin \dfrac{2 \pi t_{1}}{3}\) \(\therefore \dfrac{\pi}{6}=\dfrac{2 \pi t_{1}}{3} \Rightarrow t_{1}=\dfrac{1}{4} s\) The same will be the time to move \(12.5\;cm\) to the other side of its mean position, therefore, total time \(t = {t_1} + {t_2} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} = 0.5\;s\)
PHXI14:OSCILLATIONS
364179
If \({x=5 \sin \left(\pi t+\dfrac{\pi}{3}\right) m}\) represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are
1 \(5\;cm,2\;s\)
2 \(5\,m,2\,s\)
3 \(5\;cm,1\;s\)
4 \(5\,m,1\,s\)
Explanation:
\({x=5 \sin \left(\pi t+\dfrac{\pi}{3}\right) m}\) comparing the given equation with standard one \({x=A \sin (\omega t+\phi)}\) Amplitude \({=5 m}\) \({\omega=\pi=\dfrac{2 \pi}{T}}\) \({T=\dfrac{2 \pi}{\pi}=2 s}\)
364176
Two simple harmonic motions are represented by \({y_1} = 5\left[ {\sin 2\pi \,t + \sqrt 3 \cos 2\pi \,t} \right]\) and \({y_2} = 5\sin \left( {2\pi \,t + \frac{\pi }{4}} \right).\) The ratio of their amplitudes is
364177
Statement A : If the amplitude of SHM of a spring mass system is increased, then time period of SHM will remain constant. Statement B : If amplitude is increased, body will have to travel more distance to complete one oscillation.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both statements are correct.
4 Both Statements are incorrect.
Explanation:
Time period of oscillation of spring mass system is \(T=2 \pi \sqrt{\dfrac{m}{k}}\) which is independent of the amplitude. Thus, if the amplitude of the system is increased, then \(T\) will remain same. So option (1) is correct.
PHXI14:OSCILLATIONS
364178
A particle performs SHM with amplitude \(25\;cm\) and period \(3\;s\). The minimum time required for it to move between two points \(12.5\;cm\) on either side of the mean position is
1 \(0.6\;s\)
2 \(0.5\;s\)
3 \(0.4\;s\)
4 \(0.2\;s\)
Explanation:
Here, amplitude of particle, \(A = 25\;cm\) and time period, \(T = 3\;s\) If the particle at \(t = 0\) is at mean position, its displacement equation will be \(x=A \sin \omega t\) i.e., \(x=25 \sin \dfrac{2 \pi t}{3}\left[\because \omega=2 \pi v=\dfrac{2 \pi}{T}=\dfrac{2 \pi}{3}\right]\) If it takes time \(t_{1}\) to move a distance \(x = 12.5\;cm\) to one side of its mean position, then \(12.5=25 \sin \dfrac{2 \pi t_{1}}{3}\) or \(\dfrac{1}{2}=\sin \dfrac{2 \pi t_{1}}{3}\) or \(\sin \dfrac{\pi}{6}=\sin \dfrac{2 \pi t_{1}}{3}\) \(\therefore \dfrac{\pi}{6}=\dfrac{2 \pi t_{1}}{3} \Rightarrow t_{1}=\dfrac{1}{4} s\) The same will be the time to move \(12.5\;cm\) to the other side of its mean position, therefore, total time \(t = {t_1} + {t_2} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} = 0.5\;s\)
PHXI14:OSCILLATIONS
364179
If \({x=5 \sin \left(\pi t+\dfrac{\pi}{3}\right) m}\) represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are
1 \(5\;cm,2\;s\)
2 \(5\,m,2\,s\)
3 \(5\;cm,1\;s\)
4 \(5\,m,1\,s\)
Explanation:
\({x=5 \sin \left(\pi t+\dfrac{\pi}{3}\right) m}\) comparing the given equation with standard one \({x=A \sin (\omega t+\phi)}\) Amplitude \({=5 m}\) \({\omega=\pi=\dfrac{2 \pi}{T}}\) \({T=\dfrac{2 \pi}{\pi}=2 s}\)
364176
Two simple harmonic motions are represented by \({y_1} = 5\left[ {\sin 2\pi \,t + \sqrt 3 \cos 2\pi \,t} \right]\) and \({y_2} = 5\sin \left( {2\pi \,t + \frac{\pi }{4}} \right).\) The ratio of their amplitudes is
364177
Statement A : If the amplitude of SHM of a spring mass system is increased, then time period of SHM will remain constant. Statement B : If amplitude is increased, body will have to travel more distance to complete one oscillation.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both statements are correct.
4 Both Statements are incorrect.
Explanation:
Time period of oscillation of spring mass system is \(T=2 \pi \sqrt{\dfrac{m}{k}}\) which is independent of the amplitude. Thus, if the amplitude of the system is increased, then \(T\) will remain same. So option (1) is correct.
PHXI14:OSCILLATIONS
364178
A particle performs SHM with amplitude \(25\;cm\) and period \(3\;s\). The minimum time required for it to move between two points \(12.5\;cm\) on either side of the mean position is
1 \(0.6\;s\)
2 \(0.5\;s\)
3 \(0.4\;s\)
4 \(0.2\;s\)
Explanation:
Here, amplitude of particle, \(A = 25\;cm\) and time period, \(T = 3\;s\) If the particle at \(t = 0\) is at mean position, its displacement equation will be \(x=A \sin \omega t\) i.e., \(x=25 \sin \dfrac{2 \pi t}{3}\left[\because \omega=2 \pi v=\dfrac{2 \pi}{T}=\dfrac{2 \pi}{3}\right]\) If it takes time \(t_{1}\) to move a distance \(x = 12.5\;cm\) to one side of its mean position, then \(12.5=25 \sin \dfrac{2 \pi t_{1}}{3}\) or \(\dfrac{1}{2}=\sin \dfrac{2 \pi t_{1}}{3}\) or \(\sin \dfrac{\pi}{6}=\sin \dfrac{2 \pi t_{1}}{3}\) \(\therefore \dfrac{\pi}{6}=\dfrac{2 \pi t_{1}}{3} \Rightarrow t_{1}=\dfrac{1}{4} s\) The same will be the time to move \(12.5\;cm\) to the other side of its mean position, therefore, total time \(t = {t_1} + {t_2} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} = 0.5\;s\)
PHXI14:OSCILLATIONS
364179
If \({x=5 \sin \left(\pi t+\dfrac{\pi}{3}\right) m}\) represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are
1 \(5\;cm,2\;s\)
2 \(5\,m,2\,s\)
3 \(5\;cm,1\;s\)
4 \(5\,m,1\,s\)
Explanation:
\({x=5 \sin \left(\pi t+\dfrac{\pi}{3}\right) m}\) comparing the given equation with standard one \({x=A \sin (\omega t+\phi)}\) Amplitude \({=5 m}\) \({\omega=\pi=\dfrac{2 \pi}{T}}\) \({T=\dfrac{2 \pi}{\pi}=2 s}\)