364207
The displacement of a particle executing simple harmonic motion is given by \(y=A_{0}+A \sin \omega t+B \cos \omega t\) Then the amplitude of its oscillation is given by :
364208
Statement A : \(x=A \cos \omega t\) and \(x=A \sin \omega t\) can represent same motion with different initial positions. Statement B : If the argument of \(x=A \sin \omega t\), i.e., \(\omega t\) is increased by \(2 \pi\) radian the value of \(x\) remains same.
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:
\(x=A \cos \omega t\) and \(x=A \sin \omega t\) both represent the displacement of particle undergoing periodic motion, At \(t=0\), If \(x=A\), we can represents its displacement by \(x=A \cos \omega t\) and if \(x=0\), we can represent its displacement by \(x=A \sin \omega t\) This implies, both \(x=A \cos \omega t\) and \(x=A \sin \omega t\) represents the same motion with different initial positions. Since, \(x=A \sin \omega t\) represents a periodic function with time period \(2{\mkern 1mu} \pi \,rad\). So, if the argument of this function \(\omega t\) is increased by an integral multiple of \(2{\mkern 1mu} \pi \,rad\), the value of the function remains the same.
PHXI14:OSCILLATIONS
364209
A simple harmonic motion is represented by \(y = 5(\sin 3\pi t + \sqrt 3 \cos 3\pi t)cm\). The amplitude and time period of the motion are
1 \(5\;cm,\frac{3}{2}\;s\)
2 \(10\;cm,\frac{2}{3}\;s\)
3 \(5\;cm,\frac{2}{3}\;s\)
4 \(10\;cm,\frac{3}{2}\;s\)
Explanation:
The given equation is \(y = 5(\sin 3\pi t + \sqrt 3 \cos 3\pi t)\) \( \Rightarrow y = 2 \times 5\left( {\frac{1}{2}\sin 3\pi t + \frac{{\sqrt 3 }}{2}\cos 3\pi t} \right)\) \( = 10\left( {\cos \frac{\pi }{3}\sin 3\pi t + \sin \frac{\pi }{3}\cos 3\pi t} \right)\) \(y=10 \sin \left(3 \pi t+\dfrac{\pi}{3}\right)\) Comparing it with standard equation \(y = A\sin (\omega t + \phi )\) \( \Rightarrow \omega = 3\pi {\rm{ }}and{\rm{ }}A = 10\;cm\) \(\therefore\) Time period \(T = \frac{{2\pi }}{\omega } = \frac{2}{3}\;s\)
364207
The displacement of a particle executing simple harmonic motion is given by \(y=A_{0}+A \sin \omega t+B \cos \omega t\) Then the amplitude of its oscillation is given by :
364208
Statement A : \(x=A \cos \omega t\) and \(x=A \sin \omega t\) can represent same motion with different initial positions. Statement B : If the argument of \(x=A \sin \omega t\), i.e., \(\omega t\) is increased by \(2 \pi\) radian the value of \(x\) remains same.
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:
\(x=A \cos \omega t\) and \(x=A \sin \omega t\) both represent the displacement of particle undergoing periodic motion, At \(t=0\), If \(x=A\), we can represents its displacement by \(x=A \cos \omega t\) and if \(x=0\), we can represent its displacement by \(x=A \sin \omega t\) This implies, both \(x=A \cos \omega t\) and \(x=A \sin \omega t\) represents the same motion with different initial positions. Since, \(x=A \sin \omega t\) represents a periodic function with time period \(2{\mkern 1mu} \pi \,rad\). So, if the argument of this function \(\omega t\) is increased by an integral multiple of \(2{\mkern 1mu} \pi \,rad\), the value of the function remains the same.
PHXI14:OSCILLATIONS
364209
A simple harmonic motion is represented by \(y = 5(\sin 3\pi t + \sqrt 3 \cos 3\pi t)cm\). The amplitude and time period of the motion are
1 \(5\;cm,\frac{3}{2}\;s\)
2 \(10\;cm,\frac{2}{3}\;s\)
3 \(5\;cm,\frac{2}{3}\;s\)
4 \(10\;cm,\frac{3}{2}\;s\)
Explanation:
The given equation is \(y = 5(\sin 3\pi t + \sqrt 3 \cos 3\pi t)\) \( \Rightarrow y = 2 \times 5\left( {\frac{1}{2}\sin 3\pi t + \frac{{\sqrt 3 }}{2}\cos 3\pi t} \right)\) \( = 10\left( {\cos \frac{\pi }{3}\sin 3\pi t + \sin \frac{\pi }{3}\cos 3\pi t} \right)\) \(y=10 \sin \left(3 \pi t+\dfrac{\pi}{3}\right)\) Comparing it with standard equation \(y = A\sin (\omega t + \phi )\) \( \Rightarrow \omega = 3\pi {\rm{ }}and{\rm{ }}A = 10\;cm\) \(\therefore\) Time period \(T = \frac{{2\pi }}{\omega } = \frac{2}{3}\;s\)
364207
The displacement of a particle executing simple harmonic motion is given by \(y=A_{0}+A \sin \omega t+B \cos \omega t\) Then the amplitude of its oscillation is given by :
364208
Statement A : \(x=A \cos \omega t\) and \(x=A \sin \omega t\) can represent same motion with different initial positions. Statement B : If the argument of \(x=A \sin \omega t\), i.e., \(\omega t\) is increased by \(2 \pi\) radian the value of \(x\) remains same.
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:
\(x=A \cos \omega t\) and \(x=A \sin \omega t\) both represent the displacement of particle undergoing periodic motion, At \(t=0\), If \(x=A\), we can represents its displacement by \(x=A \cos \omega t\) and if \(x=0\), we can represent its displacement by \(x=A \sin \omega t\) This implies, both \(x=A \cos \omega t\) and \(x=A \sin \omega t\) represents the same motion with different initial positions. Since, \(x=A \sin \omega t\) represents a periodic function with time period \(2{\mkern 1mu} \pi \,rad\). So, if the argument of this function \(\omega t\) is increased by an integral multiple of \(2{\mkern 1mu} \pi \,rad\), the value of the function remains the same.
PHXI14:OSCILLATIONS
364209
A simple harmonic motion is represented by \(y = 5(\sin 3\pi t + \sqrt 3 \cos 3\pi t)cm\). The amplitude and time period of the motion are
1 \(5\;cm,\frac{3}{2}\;s\)
2 \(10\;cm,\frac{2}{3}\;s\)
3 \(5\;cm,\frac{2}{3}\;s\)
4 \(10\;cm,\frac{3}{2}\;s\)
Explanation:
The given equation is \(y = 5(\sin 3\pi t + \sqrt 3 \cos 3\pi t)\) \( \Rightarrow y = 2 \times 5\left( {\frac{1}{2}\sin 3\pi t + \frac{{\sqrt 3 }}{2}\cos 3\pi t} \right)\) \( = 10\left( {\cos \frac{\pi }{3}\sin 3\pi t + \sin \frac{\pi }{3}\cos 3\pi t} \right)\) \(y=10 \sin \left(3 \pi t+\dfrac{\pi}{3}\right)\) Comparing it with standard equation \(y = A\sin (\omega t + \phi )\) \( \Rightarrow \omega = 3\pi {\rm{ }}and{\rm{ }}A = 10\;cm\) \(\therefore\) Time period \(T = \frac{{2\pi }}{\omega } = \frac{2}{3}\;s\)
364207
The displacement of a particle executing simple harmonic motion is given by \(y=A_{0}+A \sin \omega t+B \cos \omega t\) Then the amplitude of its oscillation is given by :
364208
Statement A : \(x=A \cos \omega t\) and \(x=A \sin \omega t\) can represent same motion with different initial positions. Statement B : If the argument of \(x=A \sin \omega t\), i.e., \(\omega t\) is increased by \(2 \pi\) radian the value of \(x\) remains same.
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:
\(x=A \cos \omega t\) and \(x=A \sin \omega t\) both represent the displacement of particle undergoing periodic motion, At \(t=0\), If \(x=A\), we can represents its displacement by \(x=A \cos \omega t\) and if \(x=0\), we can represent its displacement by \(x=A \sin \omega t\) This implies, both \(x=A \cos \omega t\) and \(x=A \sin \omega t\) represents the same motion with different initial positions. Since, \(x=A \sin \omega t\) represents a periodic function with time period \(2{\mkern 1mu} \pi \,rad\). So, if the argument of this function \(\omega t\) is increased by an integral multiple of \(2{\mkern 1mu} \pi \,rad\), the value of the function remains the same.
PHXI14:OSCILLATIONS
364209
A simple harmonic motion is represented by \(y = 5(\sin 3\pi t + \sqrt 3 \cos 3\pi t)cm\). The amplitude and time period of the motion are
1 \(5\;cm,\frac{3}{2}\;s\)
2 \(10\;cm,\frac{2}{3}\;s\)
3 \(5\;cm,\frac{2}{3}\;s\)
4 \(10\;cm,\frac{3}{2}\;s\)
Explanation:
The given equation is \(y = 5(\sin 3\pi t + \sqrt 3 \cos 3\pi t)\) \( \Rightarrow y = 2 \times 5\left( {\frac{1}{2}\sin 3\pi t + \frac{{\sqrt 3 }}{2}\cos 3\pi t} \right)\) \( = 10\left( {\cos \frac{\pi }{3}\sin 3\pi t + \sin \frac{\pi }{3}\cos 3\pi t} \right)\) \(y=10 \sin \left(3 \pi t+\dfrac{\pi}{3}\right)\) Comparing it with standard equation \(y = A\sin (\omega t + \phi )\) \( \Rightarrow \omega = 3\pi {\rm{ }}and{\rm{ }}A = 10\;cm\) \(\therefore\) Time period \(T = \frac{{2\pi }}{\omega } = \frac{2}{3}\;s\)
