364219
The equation describing the motion of a simple harmonic oscillator along the \(x\)-axis is given as: \(x=A \cos (\omega t+\phi)\). If at time \(t = 0\), the oscillator is at \(x = 0\), and moving in the negative \(t\) direction, then the phase angle \(\phi\) is
364220
The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion
1 \(\pi\)
2 \(0.5\,\pi \)
3 Zero
4 \(0.707\,\pi \)
Explanation:
The displacement equation of a particle executing SHM is \(x=a \cos \omega t\) (i) velocity, \(v=\dfrac{d x}{d t}=-a \omega \sin \omega t=a \omega \cos \left(\omega t+\dfrac{\pi}{2}\right)\) (ii) and its acceleration, \(a=\dfrac{d v}{d t}=-a \omega^{2} \cos \omega t=+a \omega^{2} \cos (\omega t+\pi)\) (iii) The phase difference is \(\dfrac{\pi}{2}\)\( = 0.5{\mkern 1mu} \pi \)
PHXI14:OSCILLATIONS
364221
The phase difference between two SHM \(y_{1}=10 \sin \left(10 \pi t+\dfrac{\pi}{3}\right)\) and \(y_{2}=12 \sin \left(8 \pi t+\dfrac{\pi}{4}\right)\) at \(t = 0.5\;s\) is
364222
Two particles \(P\) and \(Q\) describe S.H.M. of same amplitude \(a\), same frequency \(f\) along the same straight line. The maximum distance between the two particles is \(a \sqrt{2}\). The phase difference between the particles is:
1 \(\dfrac{\pi}{2}\)
2 \({\rm{Zero}}\)
3 \(\dfrac{\pi}{3}\)
4 \(\dfrac{\pi}{6}\)
Explanation:
\(x_{1}=a \sin \left(\omega t+\phi_{1}\right) ; \quad x_{2}=a \sin \left(\omega t+\phi_{2}\right)\) \(\Rightarrow\left|x_{1}-x_{2}\right|=2 a \sin \left(\omega t+\dfrac{\phi_{1}+\phi_{2}}{2}\right) \cos \left(\dfrac{\phi_{1}-\phi_{2}}{2}\right)\) To maximise \(\left|x_{1}-x_{2}\right|: \sin \left(\omega t+\dfrac{\phi_{1}+\phi_{2}}{2}\right)=1\) Given that \(\left|x_{1}-x_{2}\right|_{\max }=a \sqrt{2}\) \(\begin{aligned}& \Rightarrow a \sqrt{2}=2 a \times 1 \times \cos \left(\dfrac{\phi_{1}-\phi_{2}}{2}\right) \\& \Rightarrow \dfrac{1}{\sqrt{2}}=\cos \left(\dfrac{\phi_{1}-\phi_{2}}{2}\right) \\& \Rightarrow \dfrac{\pi}{4}=\dfrac{\phi_{1}-\phi_{2}}{2} \Rightarrow \phi_{1}-\phi_{2}=\dfrac{\pi}{2}\end{aligned}\)
364219
The equation describing the motion of a simple harmonic oscillator along the \(x\)-axis is given as: \(x=A \cos (\omega t+\phi)\). If at time \(t = 0\), the oscillator is at \(x = 0\), and moving in the negative \(t\) direction, then the phase angle \(\phi\) is
364220
The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion
1 \(\pi\)
2 \(0.5\,\pi \)
3 Zero
4 \(0.707\,\pi \)
Explanation:
The displacement equation of a particle executing SHM is \(x=a \cos \omega t\) (i) velocity, \(v=\dfrac{d x}{d t}=-a \omega \sin \omega t=a \omega \cos \left(\omega t+\dfrac{\pi}{2}\right)\) (ii) and its acceleration, \(a=\dfrac{d v}{d t}=-a \omega^{2} \cos \omega t=+a \omega^{2} \cos (\omega t+\pi)\) (iii) The phase difference is \(\dfrac{\pi}{2}\)\( = 0.5{\mkern 1mu} \pi \)
PHXI14:OSCILLATIONS
364221
The phase difference between two SHM \(y_{1}=10 \sin \left(10 \pi t+\dfrac{\pi}{3}\right)\) and \(y_{2}=12 \sin \left(8 \pi t+\dfrac{\pi}{4}\right)\) at \(t = 0.5\;s\) is
364222
Two particles \(P\) and \(Q\) describe S.H.M. of same amplitude \(a\), same frequency \(f\) along the same straight line. The maximum distance between the two particles is \(a \sqrt{2}\). The phase difference between the particles is:
1 \(\dfrac{\pi}{2}\)
2 \({\rm{Zero}}\)
3 \(\dfrac{\pi}{3}\)
4 \(\dfrac{\pi}{6}\)
Explanation:
\(x_{1}=a \sin \left(\omega t+\phi_{1}\right) ; \quad x_{2}=a \sin \left(\omega t+\phi_{2}\right)\) \(\Rightarrow\left|x_{1}-x_{2}\right|=2 a \sin \left(\omega t+\dfrac{\phi_{1}+\phi_{2}}{2}\right) \cos \left(\dfrac{\phi_{1}-\phi_{2}}{2}\right)\) To maximise \(\left|x_{1}-x_{2}\right|: \sin \left(\omega t+\dfrac{\phi_{1}+\phi_{2}}{2}\right)=1\) Given that \(\left|x_{1}-x_{2}\right|_{\max }=a \sqrt{2}\) \(\begin{aligned}& \Rightarrow a \sqrt{2}=2 a \times 1 \times \cos \left(\dfrac{\phi_{1}-\phi_{2}}{2}\right) \\& \Rightarrow \dfrac{1}{\sqrt{2}}=\cos \left(\dfrac{\phi_{1}-\phi_{2}}{2}\right) \\& \Rightarrow \dfrac{\pi}{4}=\dfrac{\phi_{1}-\phi_{2}}{2} \Rightarrow \phi_{1}-\phi_{2}=\dfrac{\pi}{2}\end{aligned}\)
364219
The equation describing the motion of a simple harmonic oscillator along the \(x\)-axis is given as: \(x=A \cos (\omega t+\phi)\). If at time \(t = 0\), the oscillator is at \(x = 0\), and moving in the negative \(t\) direction, then the phase angle \(\phi\) is
364220
The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion
1 \(\pi\)
2 \(0.5\,\pi \)
3 Zero
4 \(0.707\,\pi \)
Explanation:
The displacement equation of a particle executing SHM is \(x=a \cos \omega t\) (i) velocity, \(v=\dfrac{d x}{d t}=-a \omega \sin \omega t=a \omega \cos \left(\omega t+\dfrac{\pi}{2}\right)\) (ii) and its acceleration, \(a=\dfrac{d v}{d t}=-a \omega^{2} \cos \omega t=+a \omega^{2} \cos (\omega t+\pi)\) (iii) The phase difference is \(\dfrac{\pi}{2}\)\( = 0.5{\mkern 1mu} \pi \)
PHXI14:OSCILLATIONS
364221
The phase difference between two SHM \(y_{1}=10 \sin \left(10 \pi t+\dfrac{\pi}{3}\right)\) and \(y_{2}=12 \sin \left(8 \pi t+\dfrac{\pi}{4}\right)\) at \(t = 0.5\;s\) is
364222
Two particles \(P\) and \(Q\) describe S.H.M. of same amplitude \(a\), same frequency \(f\) along the same straight line. The maximum distance between the two particles is \(a \sqrt{2}\). The phase difference between the particles is:
1 \(\dfrac{\pi}{2}\)
2 \({\rm{Zero}}\)
3 \(\dfrac{\pi}{3}\)
4 \(\dfrac{\pi}{6}\)
Explanation:
\(x_{1}=a \sin \left(\omega t+\phi_{1}\right) ; \quad x_{2}=a \sin \left(\omega t+\phi_{2}\right)\) \(\Rightarrow\left|x_{1}-x_{2}\right|=2 a \sin \left(\omega t+\dfrac{\phi_{1}+\phi_{2}}{2}\right) \cos \left(\dfrac{\phi_{1}-\phi_{2}}{2}\right)\) To maximise \(\left|x_{1}-x_{2}\right|: \sin \left(\omega t+\dfrac{\phi_{1}+\phi_{2}}{2}\right)=1\) Given that \(\left|x_{1}-x_{2}\right|_{\max }=a \sqrt{2}\) \(\begin{aligned}& \Rightarrow a \sqrt{2}=2 a \times 1 \times \cos \left(\dfrac{\phi_{1}-\phi_{2}}{2}\right) \\& \Rightarrow \dfrac{1}{\sqrt{2}}=\cos \left(\dfrac{\phi_{1}-\phi_{2}}{2}\right) \\& \Rightarrow \dfrac{\pi}{4}=\dfrac{\phi_{1}-\phi_{2}}{2} \Rightarrow \phi_{1}-\phi_{2}=\dfrac{\pi}{2}\end{aligned}\)
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PHXI14:OSCILLATIONS
364219
The equation describing the motion of a simple harmonic oscillator along the \(x\)-axis is given as: \(x=A \cos (\omega t+\phi)\). If at time \(t = 0\), the oscillator is at \(x = 0\), and moving in the negative \(t\) direction, then the phase angle \(\phi\) is
364220
The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion
1 \(\pi\)
2 \(0.5\,\pi \)
3 Zero
4 \(0.707\,\pi \)
Explanation:
The displacement equation of a particle executing SHM is \(x=a \cos \omega t\) (i) velocity, \(v=\dfrac{d x}{d t}=-a \omega \sin \omega t=a \omega \cos \left(\omega t+\dfrac{\pi}{2}\right)\) (ii) and its acceleration, \(a=\dfrac{d v}{d t}=-a \omega^{2} \cos \omega t=+a \omega^{2} \cos (\omega t+\pi)\) (iii) The phase difference is \(\dfrac{\pi}{2}\)\( = 0.5{\mkern 1mu} \pi \)
PHXI14:OSCILLATIONS
364221
The phase difference between two SHM \(y_{1}=10 \sin \left(10 \pi t+\dfrac{\pi}{3}\right)\) and \(y_{2}=12 \sin \left(8 \pi t+\dfrac{\pi}{4}\right)\) at \(t = 0.5\;s\) is
364222
Two particles \(P\) and \(Q\) describe S.H.M. of same amplitude \(a\), same frequency \(f\) along the same straight line. The maximum distance between the two particles is \(a \sqrt{2}\). The phase difference between the particles is:
1 \(\dfrac{\pi}{2}\)
2 \({\rm{Zero}}\)
3 \(\dfrac{\pi}{3}\)
4 \(\dfrac{\pi}{6}\)
Explanation:
\(x_{1}=a \sin \left(\omega t+\phi_{1}\right) ; \quad x_{2}=a \sin \left(\omega t+\phi_{2}\right)\) \(\Rightarrow\left|x_{1}-x_{2}\right|=2 a \sin \left(\omega t+\dfrac{\phi_{1}+\phi_{2}}{2}\right) \cos \left(\dfrac{\phi_{1}-\phi_{2}}{2}\right)\) To maximise \(\left|x_{1}-x_{2}\right|: \sin \left(\omega t+\dfrac{\phi_{1}+\phi_{2}}{2}\right)=1\) Given that \(\left|x_{1}-x_{2}\right|_{\max }=a \sqrt{2}\) \(\begin{aligned}& \Rightarrow a \sqrt{2}=2 a \times 1 \times \cos \left(\dfrac{\phi_{1}-\phi_{2}}{2}\right) \\& \Rightarrow \dfrac{1}{\sqrt{2}}=\cos \left(\dfrac{\phi_{1}-\phi_{2}}{2}\right) \\& \Rightarrow \dfrac{\pi}{4}=\dfrac{\phi_{1}-\phi_{2}}{2} \Rightarrow \phi_{1}-\phi_{2}=\dfrac{\pi}{2}\end{aligned}\)
