364180
The amplitude and the time period in a SHM is \(0.5\;cm\) and \(0.4\;s\) respectively. If the initial phase is \(\frac{\pi }{2}rad\), then the equation of SHM will be
364181
A particle executing S.H.M of amplitude \(4\;cm\) and \(T = 4\,\sec \). The time taken by it to move from positive extreme position to half the amplitude is
1 \(1/3\,\sec \)
2 \(1\,\sec \)
3 \(\sqrt {3/2} \,\sec \)
4 \(2/3\,\sec \)
Explanation:
Equation of motion \(y=a \cos \omega t\) \( \Rightarrow \frac{a}{2} = a\cos \omega t \Rightarrow \cos \omega t = \frac{1}{2} \Rightarrow \omega t = \frac{\pi }{3}\) \( \Rightarrow \frac{{2\pi t}}{T} = \frac{\pi }{3} \Rightarrow t = \frac{{\frac{\pi }{3} \times T}}{{2\pi }} = \frac{4}{{3 \times 2}} = \frac{2}{3}\sec \)
PHXI14:OSCILLATIONS
364182
Two particles are executing SHMs. The equations of their motions are \({y_1} = 10\sin \left( {\omega t + \frac{\pi }{4}} \right)\,\,\,\,\,{y_2} = 5\sin \left( {\omega t + \frac{{\sqrt 3 \pi }}{4}} \right)\) What is the ratio of their amplitudes.
1 \(1: 1\)
2 \(2: 1\)
3 \(1: 2\)
4 None of these
Explanation:
There is no effect of phase angle on amplitude. hence \(\dfrac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\dfrac{10}{5}=\dfrac{2}{1}\)
PHXI14:OSCILLATIONS
364183
The displacement of a particle in simple harmonic motion in one time period is
1 \(2\;A\)
2 \(A\)
3 \({\rm{Zero}}\)
4 \(4\;A\)
Explanation:
In one complete oscillation particle comes to its starting position, so displacement is zero.
PHXI14:OSCILLATIONS
364184
A particle moves with simple harmonic motion in a straight line. In first \(\tau \,s\), after starting from rest it travels a distance \(a\), and in next \(\tau \,s\) it travels \(2 a\), in same direction, then :
1 amplitude of motion is \(4 a\)
2 time period of oscillations is \(6\,\tau \)
3 amplitude of motion is \(3 a\)
4 time period of oscillations is \(8\,\tau \)
Explanation:
The displacement of the particle is \(x=A \cos \omega t\) Given that \(A(1 - \cos \omega \tau ) = a\) \(\begin{aligned}& \text { A }(1-\cos 2 \omega \tau)=3 a \\& \cos \omega \tau=\left(1-\dfrac{a}{A}\right) \\& \cos 2 \omega \tau=\left(1-\dfrac{3 a}{A}\right) \\& 2\left(1-\dfrac{a}{A}\right)^{2}-1=1-\dfrac{3 a}{A} \\& \dfrac{a}{A}=\dfrac{1}{2} \Rightarrow A=2 a \\& \cos \omega \tau=\dfrac{1}{2} \\& T=6 \tau\end{aligned}\)
364180
The amplitude and the time period in a SHM is \(0.5\;cm\) and \(0.4\;s\) respectively. If the initial phase is \(\frac{\pi }{2}rad\), then the equation of SHM will be
364181
A particle executing S.H.M of amplitude \(4\;cm\) and \(T = 4\,\sec \). The time taken by it to move from positive extreme position to half the amplitude is
1 \(1/3\,\sec \)
2 \(1\,\sec \)
3 \(\sqrt {3/2} \,\sec \)
4 \(2/3\,\sec \)
Explanation:
Equation of motion \(y=a \cos \omega t\) \( \Rightarrow \frac{a}{2} = a\cos \omega t \Rightarrow \cos \omega t = \frac{1}{2} \Rightarrow \omega t = \frac{\pi }{3}\) \( \Rightarrow \frac{{2\pi t}}{T} = \frac{\pi }{3} \Rightarrow t = \frac{{\frac{\pi }{3} \times T}}{{2\pi }} = \frac{4}{{3 \times 2}} = \frac{2}{3}\sec \)
PHXI14:OSCILLATIONS
364182
Two particles are executing SHMs. The equations of their motions are \({y_1} = 10\sin \left( {\omega t + \frac{\pi }{4}} \right)\,\,\,\,\,{y_2} = 5\sin \left( {\omega t + \frac{{\sqrt 3 \pi }}{4}} \right)\) What is the ratio of their amplitudes.
1 \(1: 1\)
2 \(2: 1\)
3 \(1: 2\)
4 None of these
Explanation:
There is no effect of phase angle on amplitude. hence \(\dfrac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\dfrac{10}{5}=\dfrac{2}{1}\)
PHXI14:OSCILLATIONS
364183
The displacement of a particle in simple harmonic motion in one time period is
1 \(2\;A\)
2 \(A\)
3 \({\rm{Zero}}\)
4 \(4\;A\)
Explanation:
In one complete oscillation particle comes to its starting position, so displacement is zero.
PHXI14:OSCILLATIONS
364184
A particle moves with simple harmonic motion in a straight line. In first \(\tau \,s\), after starting from rest it travels a distance \(a\), and in next \(\tau \,s\) it travels \(2 a\), in same direction, then :
1 amplitude of motion is \(4 a\)
2 time period of oscillations is \(6\,\tau \)
3 amplitude of motion is \(3 a\)
4 time period of oscillations is \(8\,\tau \)
Explanation:
The displacement of the particle is \(x=A \cos \omega t\) Given that \(A(1 - \cos \omega \tau ) = a\) \(\begin{aligned}& \text { A }(1-\cos 2 \omega \tau)=3 a \\& \cos \omega \tau=\left(1-\dfrac{a}{A}\right) \\& \cos 2 \omega \tau=\left(1-\dfrac{3 a}{A}\right) \\& 2\left(1-\dfrac{a}{A}\right)^{2}-1=1-\dfrac{3 a}{A} \\& \dfrac{a}{A}=\dfrac{1}{2} \Rightarrow A=2 a \\& \cos \omega \tau=\dfrac{1}{2} \\& T=6 \tau\end{aligned}\)
364180
The amplitude and the time period in a SHM is \(0.5\;cm\) and \(0.4\;s\) respectively. If the initial phase is \(\frac{\pi }{2}rad\), then the equation of SHM will be
364181
A particle executing S.H.M of amplitude \(4\;cm\) and \(T = 4\,\sec \). The time taken by it to move from positive extreme position to half the amplitude is
1 \(1/3\,\sec \)
2 \(1\,\sec \)
3 \(\sqrt {3/2} \,\sec \)
4 \(2/3\,\sec \)
Explanation:
Equation of motion \(y=a \cos \omega t\) \( \Rightarrow \frac{a}{2} = a\cos \omega t \Rightarrow \cos \omega t = \frac{1}{2} \Rightarrow \omega t = \frac{\pi }{3}\) \( \Rightarrow \frac{{2\pi t}}{T} = \frac{\pi }{3} \Rightarrow t = \frac{{\frac{\pi }{3} \times T}}{{2\pi }} = \frac{4}{{3 \times 2}} = \frac{2}{3}\sec \)
PHXI14:OSCILLATIONS
364182
Two particles are executing SHMs. The equations of their motions are \({y_1} = 10\sin \left( {\omega t + \frac{\pi }{4}} \right)\,\,\,\,\,{y_2} = 5\sin \left( {\omega t + \frac{{\sqrt 3 \pi }}{4}} \right)\) What is the ratio of their amplitudes.
1 \(1: 1\)
2 \(2: 1\)
3 \(1: 2\)
4 None of these
Explanation:
There is no effect of phase angle on amplitude. hence \(\dfrac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\dfrac{10}{5}=\dfrac{2}{1}\)
PHXI14:OSCILLATIONS
364183
The displacement of a particle in simple harmonic motion in one time period is
1 \(2\;A\)
2 \(A\)
3 \({\rm{Zero}}\)
4 \(4\;A\)
Explanation:
In one complete oscillation particle comes to its starting position, so displacement is zero.
PHXI14:OSCILLATIONS
364184
A particle moves with simple harmonic motion in a straight line. In first \(\tau \,s\), after starting from rest it travels a distance \(a\), and in next \(\tau \,s\) it travels \(2 a\), in same direction, then :
1 amplitude of motion is \(4 a\)
2 time period of oscillations is \(6\,\tau \)
3 amplitude of motion is \(3 a\)
4 time period of oscillations is \(8\,\tau \)
Explanation:
The displacement of the particle is \(x=A \cos \omega t\) Given that \(A(1 - \cos \omega \tau ) = a\) \(\begin{aligned}& \text { A }(1-\cos 2 \omega \tau)=3 a \\& \cos \omega \tau=\left(1-\dfrac{a}{A}\right) \\& \cos 2 \omega \tau=\left(1-\dfrac{3 a}{A}\right) \\& 2\left(1-\dfrac{a}{A}\right)^{2}-1=1-\dfrac{3 a}{A} \\& \dfrac{a}{A}=\dfrac{1}{2} \Rightarrow A=2 a \\& \cos \omega \tau=\dfrac{1}{2} \\& T=6 \tau\end{aligned}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXI14:OSCILLATIONS
364180
The amplitude and the time period in a SHM is \(0.5\;cm\) and \(0.4\;s\) respectively. If the initial phase is \(\frac{\pi }{2}rad\), then the equation of SHM will be
364181
A particle executing S.H.M of amplitude \(4\;cm\) and \(T = 4\,\sec \). The time taken by it to move from positive extreme position to half the amplitude is
1 \(1/3\,\sec \)
2 \(1\,\sec \)
3 \(\sqrt {3/2} \,\sec \)
4 \(2/3\,\sec \)
Explanation:
Equation of motion \(y=a \cos \omega t\) \( \Rightarrow \frac{a}{2} = a\cos \omega t \Rightarrow \cos \omega t = \frac{1}{2} \Rightarrow \omega t = \frac{\pi }{3}\) \( \Rightarrow \frac{{2\pi t}}{T} = \frac{\pi }{3} \Rightarrow t = \frac{{\frac{\pi }{3} \times T}}{{2\pi }} = \frac{4}{{3 \times 2}} = \frac{2}{3}\sec \)
PHXI14:OSCILLATIONS
364182
Two particles are executing SHMs. The equations of their motions are \({y_1} = 10\sin \left( {\omega t + \frac{\pi }{4}} \right)\,\,\,\,\,{y_2} = 5\sin \left( {\omega t + \frac{{\sqrt 3 \pi }}{4}} \right)\) What is the ratio of their amplitudes.
1 \(1: 1\)
2 \(2: 1\)
3 \(1: 2\)
4 None of these
Explanation:
There is no effect of phase angle on amplitude. hence \(\dfrac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\dfrac{10}{5}=\dfrac{2}{1}\)
PHXI14:OSCILLATIONS
364183
The displacement of a particle in simple harmonic motion in one time period is
1 \(2\;A\)
2 \(A\)
3 \({\rm{Zero}}\)
4 \(4\;A\)
Explanation:
In one complete oscillation particle comes to its starting position, so displacement is zero.
PHXI14:OSCILLATIONS
364184
A particle moves with simple harmonic motion in a straight line. In first \(\tau \,s\), after starting from rest it travels a distance \(a\), and in next \(\tau \,s\) it travels \(2 a\), in same direction, then :
1 amplitude of motion is \(4 a\)
2 time period of oscillations is \(6\,\tau \)
3 amplitude of motion is \(3 a\)
4 time period of oscillations is \(8\,\tau \)
Explanation:
The displacement of the particle is \(x=A \cos \omega t\) Given that \(A(1 - \cos \omega \tau ) = a\) \(\begin{aligned}& \text { A }(1-\cos 2 \omega \tau)=3 a \\& \cos \omega \tau=\left(1-\dfrac{a}{A}\right) \\& \cos 2 \omega \tau=\left(1-\dfrac{3 a}{A}\right) \\& 2\left(1-\dfrac{a}{A}\right)^{2}-1=1-\dfrac{3 a}{A} \\& \dfrac{a}{A}=\dfrac{1}{2} \Rightarrow A=2 a \\& \cos \omega \tau=\dfrac{1}{2} \\& T=6 \tau\end{aligned}\)
364180
The amplitude and the time period in a SHM is \(0.5\;cm\) and \(0.4\;s\) respectively. If the initial phase is \(\frac{\pi }{2}rad\), then the equation of SHM will be
364181
A particle executing S.H.M of amplitude \(4\;cm\) and \(T = 4\,\sec \). The time taken by it to move from positive extreme position to half the amplitude is
1 \(1/3\,\sec \)
2 \(1\,\sec \)
3 \(\sqrt {3/2} \,\sec \)
4 \(2/3\,\sec \)
Explanation:
Equation of motion \(y=a \cos \omega t\) \( \Rightarrow \frac{a}{2} = a\cos \omega t \Rightarrow \cos \omega t = \frac{1}{2} \Rightarrow \omega t = \frac{\pi }{3}\) \( \Rightarrow \frac{{2\pi t}}{T} = \frac{\pi }{3} \Rightarrow t = \frac{{\frac{\pi }{3} \times T}}{{2\pi }} = \frac{4}{{3 \times 2}} = \frac{2}{3}\sec \)
PHXI14:OSCILLATIONS
364182
Two particles are executing SHMs. The equations of their motions are \({y_1} = 10\sin \left( {\omega t + \frac{\pi }{4}} \right)\,\,\,\,\,{y_2} = 5\sin \left( {\omega t + \frac{{\sqrt 3 \pi }}{4}} \right)\) What is the ratio of their amplitudes.
1 \(1: 1\)
2 \(2: 1\)
3 \(1: 2\)
4 None of these
Explanation:
There is no effect of phase angle on amplitude. hence \(\dfrac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\dfrac{10}{5}=\dfrac{2}{1}\)
PHXI14:OSCILLATIONS
364183
The displacement of a particle in simple harmonic motion in one time period is
1 \(2\;A\)
2 \(A\)
3 \({\rm{Zero}}\)
4 \(4\;A\)
Explanation:
In one complete oscillation particle comes to its starting position, so displacement is zero.
PHXI14:OSCILLATIONS
364184
A particle moves with simple harmonic motion in a straight line. In first \(\tau \,s\), after starting from rest it travels a distance \(a\), and in next \(\tau \,s\) it travels \(2 a\), in same direction, then :
1 amplitude of motion is \(4 a\)
2 time period of oscillations is \(6\,\tau \)
3 amplitude of motion is \(3 a\)
4 time period of oscillations is \(8\,\tau \)
Explanation:
The displacement of the particle is \(x=A \cos \omega t\) Given that \(A(1 - \cos \omega \tau ) = a\) \(\begin{aligned}& \text { A }(1-\cos 2 \omega \tau)=3 a \\& \cos \omega \tau=\left(1-\dfrac{a}{A}\right) \\& \cos 2 \omega \tau=\left(1-\dfrac{3 a}{A}\right) \\& 2\left(1-\dfrac{a}{A}\right)^{2}-1=1-\dfrac{3 a}{A} \\& \dfrac{a}{A}=\dfrac{1}{2} \Rightarrow A=2 a \\& \cos \omega \tau=\dfrac{1}{2} \\& T=6 \tau\end{aligned}\)