364172
Which of the following functions represents a simple harmonic oscillation?
1 \(\sin \omega t+\sin (3 \omega t)\)
2 \(\sin \omega t-\cos \omega t\)
3 \(\sin \omega t-\sin 2 \omega t\)
4 \(\sin \omega t+\sin 2 \omega t\)
Explanation:
One of the condition for SHM is that restoring force \((F)\) and hence acceleration \((a)\) should be proportional to displacement \((y)\). Let, \(\quad y=\sin \omega t-\cos \omega t\) \(\begin{gathered}\dfrac{d y}{d t}=\omega \cos \omega t+\omega \sin \omega t \\\dfrac{d^{2} y}{d t^{2}}=-\omega^{2} \cos \omega t+\omega^{2} \cos \omega t \\\text { or } \quad a=-\omega^{2}(\sin \omega t-\cos \omega t) \\a=-\omega^{2} y \Rightarrow a \propto-y\end{gathered}\) This satisfies the condition of SHM, other equations do not satisfy this condition.
PHXI14:OSCILLATIONS
364173
The motion of a particle executing \(S H M\) in one dimension is described by \(x=-0.3 \sin \left(t+\dfrac{\pi}{4}\right)\), where \(x\) is in metre and \(t\) in second. The frequency of oscillation in \(\mathrm{Hz}\) is
1 3
2 \(\dfrac{1}{2 \pi}\)
3 \(\dfrac{\pi}{2}\)
4 \(\dfrac{1}{\pi}\)
Explanation:
We have, \(x=-0.3 \sin \left(t+\dfrac{\pi}{4}\right)\) Comparing with the general equation \(x=x_{0} \sin (\omega t+\phi)\) where, \(x_{0}=\) maximum displacement So, \(\quad x_{0}=0.3, \quad \omega=1, \quad \phi=\dfrac{\pi}{4}\) Hence, \(2 \pi f=1 \Rightarrow f=1 / 2 \pi\)
PHXI14:OSCILLATIONS
364174
The displacement of a particle is represented by the equation \(y=\sin ^{3} \omega t\). The motion is
1 Non-periodic
2 Periodic but not simple harmonic
3 Simple harmonic motion with period \(\pi / \omega\)
4 Simple harmonic motion with period \(2 \pi / \omega\)
Explanation:
Given, \(y=\sin ^{3} \omega t=\dfrac{1}{4}[3 \sin \omega t-\sin 3 \omega t]\) As this motion cannot be represented by single sin or cos function, hence it is not SHM. As this motion involves sine and cosine functions, hence it is periodic motion.
NCERT Exemplar
PHXI14:OSCILLATIONS
364175
A body oscillates with SHM according to the equation \(x=5 \cos \left(2 \pi t+\dfrac{\pi}{4}\right)\). Its instantaneous displacement at \(t=1 \mathrm{sec}\) is
1 \(\dfrac{\sqrt{2}}{5}\)
2 \(\dfrac{1}{\sqrt{3}}\)
3 \(\dfrac{1}{\sqrt{2}}\)
4 \(\dfrac{5}{\sqrt{2}}\)
Explanation:
For instantaneous displacement Put \(t=1 \mathrm{sec}\) \(x=5 \cos \left[2 \pi+\dfrac{\pi}{4}\right]\) \(=5 \cos \dfrac{\pi}{4}=\dfrac{5}{\sqrt{2}}\).
364172
Which of the following functions represents a simple harmonic oscillation?
1 \(\sin \omega t+\sin (3 \omega t)\)
2 \(\sin \omega t-\cos \omega t\)
3 \(\sin \omega t-\sin 2 \omega t\)
4 \(\sin \omega t+\sin 2 \omega t\)
Explanation:
One of the condition for SHM is that restoring force \((F)\) and hence acceleration \((a)\) should be proportional to displacement \((y)\). Let, \(\quad y=\sin \omega t-\cos \omega t\) \(\begin{gathered}\dfrac{d y}{d t}=\omega \cos \omega t+\omega \sin \omega t \\\dfrac{d^{2} y}{d t^{2}}=-\omega^{2} \cos \omega t+\omega^{2} \cos \omega t \\\text { or } \quad a=-\omega^{2}(\sin \omega t-\cos \omega t) \\a=-\omega^{2} y \Rightarrow a \propto-y\end{gathered}\) This satisfies the condition of SHM, other equations do not satisfy this condition.
PHXI14:OSCILLATIONS
364173
The motion of a particle executing \(S H M\) in one dimension is described by \(x=-0.3 \sin \left(t+\dfrac{\pi}{4}\right)\), where \(x\) is in metre and \(t\) in second. The frequency of oscillation in \(\mathrm{Hz}\) is
1 3
2 \(\dfrac{1}{2 \pi}\)
3 \(\dfrac{\pi}{2}\)
4 \(\dfrac{1}{\pi}\)
Explanation:
We have, \(x=-0.3 \sin \left(t+\dfrac{\pi}{4}\right)\) Comparing with the general equation \(x=x_{0} \sin (\omega t+\phi)\) where, \(x_{0}=\) maximum displacement So, \(\quad x_{0}=0.3, \quad \omega=1, \quad \phi=\dfrac{\pi}{4}\) Hence, \(2 \pi f=1 \Rightarrow f=1 / 2 \pi\)
PHXI14:OSCILLATIONS
364174
The displacement of a particle is represented by the equation \(y=\sin ^{3} \omega t\). The motion is
1 Non-periodic
2 Periodic but not simple harmonic
3 Simple harmonic motion with period \(\pi / \omega\)
4 Simple harmonic motion with period \(2 \pi / \omega\)
Explanation:
Given, \(y=\sin ^{3} \omega t=\dfrac{1}{4}[3 \sin \omega t-\sin 3 \omega t]\) As this motion cannot be represented by single sin or cos function, hence it is not SHM. As this motion involves sine and cosine functions, hence it is periodic motion.
NCERT Exemplar
PHXI14:OSCILLATIONS
364175
A body oscillates with SHM according to the equation \(x=5 \cos \left(2 \pi t+\dfrac{\pi}{4}\right)\). Its instantaneous displacement at \(t=1 \mathrm{sec}\) is
1 \(\dfrac{\sqrt{2}}{5}\)
2 \(\dfrac{1}{\sqrt{3}}\)
3 \(\dfrac{1}{\sqrt{2}}\)
4 \(\dfrac{5}{\sqrt{2}}\)
Explanation:
For instantaneous displacement Put \(t=1 \mathrm{sec}\) \(x=5 \cos \left[2 \pi+\dfrac{\pi}{4}\right]\) \(=5 \cos \dfrac{\pi}{4}=\dfrac{5}{\sqrt{2}}\).
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PHXI14:OSCILLATIONS
364172
Which of the following functions represents a simple harmonic oscillation?
1 \(\sin \omega t+\sin (3 \omega t)\)
2 \(\sin \omega t-\cos \omega t\)
3 \(\sin \omega t-\sin 2 \omega t\)
4 \(\sin \omega t+\sin 2 \omega t\)
Explanation:
One of the condition for SHM is that restoring force \((F)\) and hence acceleration \((a)\) should be proportional to displacement \((y)\). Let, \(\quad y=\sin \omega t-\cos \omega t\) \(\begin{gathered}\dfrac{d y}{d t}=\omega \cos \omega t+\omega \sin \omega t \\\dfrac{d^{2} y}{d t^{2}}=-\omega^{2} \cos \omega t+\omega^{2} \cos \omega t \\\text { or } \quad a=-\omega^{2}(\sin \omega t-\cos \omega t) \\a=-\omega^{2} y \Rightarrow a \propto-y\end{gathered}\) This satisfies the condition of SHM, other equations do not satisfy this condition.
PHXI14:OSCILLATIONS
364173
The motion of a particle executing \(S H M\) in one dimension is described by \(x=-0.3 \sin \left(t+\dfrac{\pi}{4}\right)\), where \(x\) is in metre and \(t\) in second. The frequency of oscillation in \(\mathrm{Hz}\) is
1 3
2 \(\dfrac{1}{2 \pi}\)
3 \(\dfrac{\pi}{2}\)
4 \(\dfrac{1}{\pi}\)
Explanation:
We have, \(x=-0.3 \sin \left(t+\dfrac{\pi}{4}\right)\) Comparing with the general equation \(x=x_{0} \sin (\omega t+\phi)\) where, \(x_{0}=\) maximum displacement So, \(\quad x_{0}=0.3, \quad \omega=1, \quad \phi=\dfrac{\pi}{4}\) Hence, \(2 \pi f=1 \Rightarrow f=1 / 2 \pi\)
PHXI14:OSCILLATIONS
364174
The displacement of a particle is represented by the equation \(y=\sin ^{3} \omega t\). The motion is
1 Non-periodic
2 Periodic but not simple harmonic
3 Simple harmonic motion with period \(\pi / \omega\)
4 Simple harmonic motion with period \(2 \pi / \omega\)
Explanation:
Given, \(y=\sin ^{3} \omega t=\dfrac{1}{4}[3 \sin \omega t-\sin 3 \omega t]\) As this motion cannot be represented by single sin or cos function, hence it is not SHM. As this motion involves sine and cosine functions, hence it is periodic motion.
NCERT Exemplar
PHXI14:OSCILLATIONS
364175
A body oscillates with SHM according to the equation \(x=5 \cos \left(2 \pi t+\dfrac{\pi}{4}\right)\). Its instantaneous displacement at \(t=1 \mathrm{sec}\) is
1 \(\dfrac{\sqrt{2}}{5}\)
2 \(\dfrac{1}{\sqrt{3}}\)
3 \(\dfrac{1}{\sqrt{2}}\)
4 \(\dfrac{5}{\sqrt{2}}\)
Explanation:
For instantaneous displacement Put \(t=1 \mathrm{sec}\) \(x=5 \cos \left[2 \pi+\dfrac{\pi}{4}\right]\) \(=5 \cos \dfrac{\pi}{4}=\dfrac{5}{\sqrt{2}}\).
364172
Which of the following functions represents a simple harmonic oscillation?
1 \(\sin \omega t+\sin (3 \omega t)\)
2 \(\sin \omega t-\cos \omega t\)
3 \(\sin \omega t-\sin 2 \omega t\)
4 \(\sin \omega t+\sin 2 \omega t\)
Explanation:
One of the condition for SHM is that restoring force \((F)\) and hence acceleration \((a)\) should be proportional to displacement \((y)\). Let, \(\quad y=\sin \omega t-\cos \omega t\) \(\begin{gathered}\dfrac{d y}{d t}=\omega \cos \omega t+\omega \sin \omega t \\\dfrac{d^{2} y}{d t^{2}}=-\omega^{2} \cos \omega t+\omega^{2} \cos \omega t \\\text { or } \quad a=-\omega^{2}(\sin \omega t-\cos \omega t) \\a=-\omega^{2} y \Rightarrow a \propto-y\end{gathered}\) This satisfies the condition of SHM, other equations do not satisfy this condition.
PHXI14:OSCILLATIONS
364173
The motion of a particle executing \(S H M\) in one dimension is described by \(x=-0.3 \sin \left(t+\dfrac{\pi}{4}\right)\), where \(x\) is in metre and \(t\) in second. The frequency of oscillation in \(\mathrm{Hz}\) is
1 3
2 \(\dfrac{1}{2 \pi}\)
3 \(\dfrac{\pi}{2}\)
4 \(\dfrac{1}{\pi}\)
Explanation:
We have, \(x=-0.3 \sin \left(t+\dfrac{\pi}{4}\right)\) Comparing with the general equation \(x=x_{0} \sin (\omega t+\phi)\) where, \(x_{0}=\) maximum displacement So, \(\quad x_{0}=0.3, \quad \omega=1, \quad \phi=\dfrac{\pi}{4}\) Hence, \(2 \pi f=1 \Rightarrow f=1 / 2 \pi\)
PHXI14:OSCILLATIONS
364174
The displacement of a particle is represented by the equation \(y=\sin ^{3} \omega t\). The motion is
1 Non-periodic
2 Periodic but not simple harmonic
3 Simple harmonic motion with period \(\pi / \omega\)
4 Simple harmonic motion with period \(2 \pi / \omega\)
Explanation:
Given, \(y=\sin ^{3} \omega t=\dfrac{1}{4}[3 \sin \omega t-\sin 3 \omega t]\) As this motion cannot be represented by single sin or cos function, hence it is not SHM. As this motion involves sine and cosine functions, hence it is periodic motion.
NCERT Exemplar
PHXI14:OSCILLATIONS
364175
A body oscillates with SHM according to the equation \(x=5 \cos \left(2 \pi t+\dfrac{\pi}{4}\right)\). Its instantaneous displacement at \(t=1 \mathrm{sec}\) is
1 \(\dfrac{\sqrt{2}}{5}\)
2 \(\dfrac{1}{\sqrt{3}}\)
3 \(\dfrac{1}{\sqrt{2}}\)
4 \(\dfrac{5}{\sqrt{2}}\)
Explanation:
For instantaneous displacement Put \(t=1 \mathrm{sec}\) \(x=5 \cos \left[2 \pi+\dfrac{\pi}{4}\right]\) \(=5 \cos \dfrac{\pi}{4}=\dfrac{5}{\sqrt{2}}\).