364198
Out of the following functions representing motion of a particle which represents SHM A. \(y=\sin ^{3} \omega t\) B. \(y=\sin \omega t-\cos \omega t\) C. \(y=1+\omega t+\omega^{2} t^{2}\) D. \(y=5 \cos \left(\dfrac{3 \pi}{4}-3 \omega t\right)\)
1 Only (B)
2 Only (A) and (B)
3 Only (B) and (D)
4 Only (C) does not represent SHM
Explanation:
The function \(y=5 \cos \left(\dfrac{3 \pi}{4}-3 \omega t\right)\) represents equation of SHM with phase constant \(\dfrac{3 \pi}{4}\) \(\begin{aligned}& y=\sin \omega t-\cos \omega t \\& \dfrac{d y}{d t}=\omega \cos \omega t+\omega \sin \omega t \\& \dfrac{d^{2} y}{d t^{2}}=-\omega^{2} \sin \omega t+\omega^{2} \cos \omega t \\& =-\omega^{2}(\sin \omega t-\cos \omega t) \\& \dfrac{d^{2} y}{d t^{2}}=-\omega^{2} y\end{aligned}\) The other two functions do not obey \(\dfrac{d^{2} y}{d t^{2}}=-\omega^{2} y\)
PHXI14:OSCILLATIONS
364199
A system executing SHM must posseses
1 Inertia only
2 Restoration force
3 Both restoring force and inertia
4 Only external force.
Explanation:
Conceptual Question
PHXI14:OSCILLATIONS
364200
A particle moves according to the equation \(x=a \cos \left(\dfrac{\pi t}{2}\right)\). The distance covered by it in the time interval between \(t=0\) to \(t=3\) is
1 \(2 a\)
2 \(3 a\)
3 \(8 a\)
4 \(a\)
Explanation:
\(x=a \cos \dfrac{\pi t}{2} \Rightarrow \omega=\dfrac{\pi}{2}\) \(\Rightarrow T=\dfrac{2 \pi}{\omega}=4 \mathrm{sec}\) also at \(t=0, x=a \Rightarrow\) particle is at extreme position. \(\Rightarrow\) The distance covered in \(1 \mathrm{sec}\) is ' \(a\) ' \(\Rightarrow\) The required distance is ' \(3 a\) '.
PHXI14:OSCILLATIONS
364201
The displacement of a particle varies according to the relation \({x=4(\cos \pi t+\sin \pi t)}\). The amplitude of the particle is
1 \({4 \sqrt{2}}\)
2 4
3 \({\dfrac{4}{\sqrt{2}}}\)
4 8
Explanation:
\({ y=4(\cos \pi t+\sin \pi t)}\) \({=4 \sqrt{2}\left(\dfrac{1}{\sqrt{2}} \cos \pi t+\dfrac{1}{\sqrt{2}} \sin \pi t\right)}\) \({=4 \sqrt{2}\left(\sin \dfrac{\pi}{4} \cos \pi t+\cos \dfrac{\pi}{4} \sin \pi t\right)}\) \({y=4 \sqrt{2} \sin \left(\dfrac{\pi}{4}+\pi t\right)}\) Comparing this equation with standard equation, we get Amplitude, \({A=4 \sqrt{2}}\) So correct option is (1)
364198
Out of the following functions representing motion of a particle which represents SHM A. \(y=\sin ^{3} \omega t\) B. \(y=\sin \omega t-\cos \omega t\) C. \(y=1+\omega t+\omega^{2} t^{2}\) D. \(y=5 \cos \left(\dfrac{3 \pi}{4}-3 \omega t\right)\)
1 Only (B)
2 Only (A) and (B)
3 Only (B) and (D)
4 Only (C) does not represent SHM
Explanation:
The function \(y=5 \cos \left(\dfrac{3 \pi}{4}-3 \omega t\right)\) represents equation of SHM with phase constant \(\dfrac{3 \pi}{4}\) \(\begin{aligned}& y=\sin \omega t-\cos \omega t \\& \dfrac{d y}{d t}=\omega \cos \omega t+\omega \sin \omega t \\& \dfrac{d^{2} y}{d t^{2}}=-\omega^{2} \sin \omega t+\omega^{2} \cos \omega t \\& =-\omega^{2}(\sin \omega t-\cos \omega t) \\& \dfrac{d^{2} y}{d t^{2}}=-\omega^{2} y\end{aligned}\) The other two functions do not obey \(\dfrac{d^{2} y}{d t^{2}}=-\omega^{2} y\)
PHXI14:OSCILLATIONS
364199
A system executing SHM must posseses
1 Inertia only
2 Restoration force
3 Both restoring force and inertia
4 Only external force.
Explanation:
Conceptual Question
PHXI14:OSCILLATIONS
364200
A particle moves according to the equation \(x=a \cos \left(\dfrac{\pi t}{2}\right)\). The distance covered by it in the time interval between \(t=0\) to \(t=3\) is
1 \(2 a\)
2 \(3 a\)
3 \(8 a\)
4 \(a\)
Explanation:
\(x=a \cos \dfrac{\pi t}{2} \Rightarrow \omega=\dfrac{\pi}{2}\) \(\Rightarrow T=\dfrac{2 \pi}{\omega}=4 \mathrm{sec}\) also at \(t=0, x=a \Rightarrow\) particle is at extreme position. \(\Rightarrow\) The distance covered in \(1 \mathrm{sec}\) is ' \(a\) ' \(\Rightarrow\) The required distance is ' \(3 a\) '.
PHXI14:OSCILLATIONS
364201
The displacement of a particle varies according to the relation \({x=4(\cos \pi t+\sin \pi t)}\). The amplitude of the particle is
1 \({4 \sqrt{2}}\)
2 4
3 \({\dfrac{4}{\sqrt{2}}}\)
4 8
Explanation:
\({ y=4(\cos \pi t+\sin \pi t)}\) \({=4 \sqrt{2}\left(\dfrac{1}{\sqrt{2}} \cos \pi t+\dfrac{1}{\sqrt{2}} \sin \pi t\right)}\) \({=4 \sqrt{2}\left(\sin \dfrac{\pi}{4} \cos \pi t+\cos \dfrac{\pi}{4} \sin \pi t\right)}\) \({y=4 \sqrt{2} \sin \left(\dfrac{\pi}{4}+\pi t\right)}\) Comparing this equation with standard equation, we get Amplitude, \({A=4 \sqrt{2}}\) So correct option is (1)
364198
Out of the following functions representing motion of a particle which represents SHM A. \(y=\sin ^{3} \omega t\) B. \(y=\sin \omega t-\cos \omega t\) C. \(y=1+\omega t+\omega^{2} t^{2}\) D. \(y=5 \cos \left(\dfrac{3 \pi}{4}-3 \omega t\right)\)
1 Only (B)
2 Only (A) and (B)
3 Only (B) and (D)
4 Only (C) does not represent SHM
Explanation:
The function \(y=5 \cos \left(\dfrac{3 \pi}{4}-3 \omega t\right)\) represents equation of SHM with phase constant \(\dfrac{3 \pi}{4}\) \(\begin{aligned}& y=\sin \omega t-\cos \omega t \\& \dfrac{d y}{d t}=\omega \cos \omega t+\omega \sin \omega t \\& \dfrac{d^{2} y}{d t^{2}}=-\omega^{2} \sin \omega t+\omega^{2} \cos \omega t \\& =-\omega^{2}(\sin \omega t-\cos \omega t) \\& \dfrac{d^{2} y}{d t^{2}}=-\omega^{2} y\end{aligned}\) The other two functions do not obey \(\dfrac{d^{2} y}{d t^{2}}=-\omega^{2} y\)
PHXI14:OSCILLATIONS
364199
A system executing SHM must posseses
1 Inertia only
2 Restoration force
3 Both restoring force and inertia
4 Only external force.
Explanation:
Conceptual Question
PHXI14:OSCILLATIONS
364200
A particle moves according to the equation \(x=a \cos \left(\dfrac{\pi t}{2}\right)\). The distance covered by it in the time interval between \(t=0\) to \(t=3\) is
1 \(2 a\)
2 \(3 a\)
3 \(8 a\)
4 \(a\)
Explanation:
\(x=a \cos \dfrac{\pi t}{2} \Rightarrow \omega=\dfrac{\pi}{2}\) \(\Rightarrow T=\dfrac{2 \pi}{\omega}=4 \mathrm{sec}\) also at \(t=0, x=a \Rightarrow\) particle is at extreme position. \(\Rightarrow\) The distance covered in \(1 \mathrm{sec}\) is ' \(a\) ' \(\Rightarrow\) The required distance is ' \(3 a\) '.
PHXI14:OSCILLATIONS
364201
The displacement of a particle varies according to the relation \({x=4(\cos \pi t+\sin \pi t)}\). The amplitude of the particle is
1 \({4 \sqrt{2}}\)
2 4
3 \({\dfrac{4}{\sqrt{2}}}\)
4 8
Explanation:
\({ y=4(\cos \pi t+\sin \pi t)}\) \({=4 \sqrt{2}\left(\dfrac{1}{\sqrt{2}} \cos \pi t+\dfrac{1}{\sqrt{2}} \sin \pi t\right)}\) \({=4 \sqrt{2}\left(\sin \dfrac{\pi}{4} \cos \pi t+\cos \dfrac{\pi}{4} \sin \pi t\right)}\) \({y=4 \sqrt{2} \sin \left(\dfrac{\pi}{4}+\pi t\right)}\) Comparing this equation with standard equation, we get Amplitude, \({A=4 \sqrt{2}}\) So correct option is (1)
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PHXI14:OSCILLATIONS
364198
Out of the following functions representing motion of a particle which represents SHM A. \(y=\sin ^{3} \omega t\) B. \(y=\sin \omega t-\cos \omega t\) C. \(y=1+\omega t+\omega^{2} t^{2}\) D. \(y=5 \cos \left(\dfrac{3 \pi}{4}-3 \omega t\right)\)
1 Only (B)
2 Only (A) and (B)
3 Only (B) and (D)
4 Only (C) does not represent SHM
Explanation:
The function \(y=5 \cos \left(\dfrac{3 \pi}{4}-3 \omega t\right)\) represents equation of SHM with phase constant \(\dfrac{3 \pi}{4}\) \(\begin{aligned}& y=\sin \omega t-\cos \omega t \\& \dfrac{d y}{d t}=\omega \cos \omega t+\omega \sin \omega t \\& \dfrac{d^{2} y}{d t^{2}}=-\omega^{2} \sin \omega t+\omega^{2} \cos \omega t \\& =-\omega^{2}(\sin \omega t-\cos \omega t) \\& \dfrac{d^{2} y}{d t^{2}}=-\omega^{2} y\end{aligned}\) The other two functions do not obey \(\dfrac{d^{2} y}{d t^{2}}=-\omega^{2} y\)
PHXI14:OSCILLATIONS
364199
A system executing SHM must posseses
1 Inertia only
2 Restoration force
3 Both restoring force and inertia
4 Only external force.
Explanation:
Conceptual Question
PHXI14:OSCILLATIONS
364200
A particle moves according to the equation \(x=a \cos \left(\dfrac{\pi t}{2}\right)\). The distance covered by it in the time interval between \(t=0\) to \(t=3\) is
1 \(2 a\)
2 \(3 a\)
3 \(8 a\)
4 \(a\)
Explanation:
\(x=a \cos \dfrac{\pi t}{2} \Rightarrow \omega=\dfrac{\pi}{2}\) \(\Rightarrow T=\dfrac{2 \pi}{\omega}=4 \mathrm{sec}\) also at \(t=0, x=a \Rightarrow\) particle is at extreme position. \(\Rightarrow\) The distance covered in \(1 \mathrm{sec}\) is ' \(a\) ' \(\Rightarrow\) The required distance is ' \(3 a\) '.
PHXI14:OSCILLATIONS
364201
The displacement of a particle varies according to the relation \({x=4(\cos \pi t+\sin \pi t)}\). The amplitude of the particle is
1 \({4 \sqrt{2}}\)
2 4
3 \({\dfrac{4}{\sqrt{2}}}\)
4 8
Explanation:
\({ y=4(\cos \pi t+\sin \pi t)}\) \({=4 \sqrt{2}\left(\dfrac{1}{\sqrt{2}} \cos \pi t+\dfrac{1}{\sqrt{2}} \sin \pi t\right)}\) \({=4 \sqrt{2}\left(\sin \dfrac{\pi}{4} \cos \pi t+\cos \dfrac{\pi}{4} \sin \pi t\right)}\) \({y=4 \sqrt{2} \sin \left(\dfrac{\pi}{4}+\pi t\right)}\) Comparing this equation with standard equation, we get Amplitude, \({A=4 \sqrt{2}}\) So correct option is (1)
