364292
A particle moves such that its acceleration \(a\) is given by \(a=-b x\), where \(x\) is the displacement from equilibrium position and \(b\) is constant. The period of oscillation is
1 \(\dfrac{2 \pi}{\sqrt{b}}\)
2 \(2 \pi \sqrt{b}\)
3 \(2 \sqrt{\dfrac{\pi}{b}}\)
4 \(\dfrac{2 \pi}{b}\)
Explanation:
Given that \(a=-b x\) Comparing \(a=-\omega^{2} x \Rightarrow \omega^{2}=b \Rightarrow \omega=\sqrt{b}\) \( \Rightarrow \frac{{2\pi }}{T} = \sqrt b \Rightarrow T = \frac{{2\pi }}{{\sqrt b }}\)
PHXI14:OSCILLATIONS
364293
Assertion : In simple harmonic motion, the motion is to and fro and periodic. Reason : Velocity of the particle \((v)=\omega \sqrt{A^{2}-x^{2}}\) (where \(x\) is the displacement and \(A\) is amplitude)
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
A particle is said to execute simple harmonic motion, if it moves to and fro about a fixed point under the action of a restoring force which is directly proportional to its displacement from the fixed point and is always directed towards the fixed point. \(x=A \cos (\omega t+\phi)\) \(v=-A \omega \sin (\omega t+\phi)\) \(\Rightarrow v=\omega \sqrt{A^{2}-x^{2}}\) So, correct option is (2).
PHXI14:OSCILLATIONS
364294
The relation between acceleration and displacement of four particles are given below: Which, one of the particle is executing simple harmonic motion?
1 \(a_{x}=+2 x\)
2 \(a_{x}=+2 x^{2}\)
3 \(a_{x}=-2 x^{2}\)
4 \(a_{x}=-2 x\)
Explanation:
For motion to be S.H.M. acceleration of the particle must be proportional to negative of displacement. i.e., \(a \propto-(y\) or \(x)\) We should be clear that \(y\) has to be linear. Correct option is (4).
PHXI14:OSCILLATIONS
364295
The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is \(10\;{s^{ - 1}}\). At, \(t = 0\) the displacement is \(5\;m\). What is the maximum acceleration? The initial phase is \(\dfrac{\pi}{4}\).
1 \(500\;m/{s^2}\)
2 \(750\sqrt 2 \;m/{s^2}\)
3 \(750\;m/{s^2}\)
4 \(500\sqrt 2 \;m/{s^2}\)
Explanation:
Let \(a_{o}\) and \(v_{o}\) are maximum acceleration and velocity respectively. Given that \(\dfrac{a_{o}}{v_{o}}=10=\dfrac{\omega^{2} A}{\omega A}=\omega\) \(\omega=10 s^{-1}, \phi=\dfrac{\pi}{4} \& x=5 m\) Let \(\quad x=A \sin \left(\omega t+\dfrac{\pi}{4}\right)\) \(\begin{aligned}& 5=A \sin \left(0+\dfrac{\pi}{4}\right) \Rightarrow A=5 \\& a_{o}=\omega^{2} A=500 \sqrt{2}\end{aligned}\)
364292
A particle moves such that its acceleration \(a\) is given by \(a=-b x\), where \(x\) is the displacement from equilibrium position and \(b\) is constant. The period of oscillation is
1 \(\dfrac{2 \pi}{\sqrt{b}}\)
2 \(2 \pi \sqrt{b}\)
3 \(2 \sqrt{\dfrac{\pi}{b}}\)
4 \(\dfrac{2 \pi}{b}\)
Explanation:
Given that \(a=-b x\) Comparing \(a=-\omega^{2} x \Rightarrow \omega^{2}=b \Rightarrow \omega=\sqrt{b}\) \( \Rightarrow \frac{{2\pi }}{T} = \sqrt b \Rightarrow T = \frac{{2\pi }}{{\sqrt b }}\)
PHXI14:OSCILLATIONS
364293
Assertion : In simple harmonic motion, the motion is to and fro and periodic. Reason : Velocity of the particle \((v)=\omega \sqrt{A^{2}-x^{2}}\) (where \(x\) is the displacement and \(A\) is amplitude)
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
A particle is said to execute simple harmonic motion, if it moves to and fro about a fixed point under the action of a restoring force which is directly proportional to its displacement from the fixed point and is always directed towards the fixed point. \(x=A \cos (\omega t+\phi)\) \(v=-A \omega \sin (\omega t+\phi)\) \(\Rightarrow v=\omega \sqrt{A^{2}-x^{2}}\) So, correct option is (2).
PHXI14:OSCILLATIONS
364294
The relation between acceleration and displacement of four particles are given below: Which, one of the particle is executing simple harmonic motion?
1 \(a_{x}=+2 x\)
2 \(a_{x}=+2 x^{2}\)
3 \(a_{x}=-2 x^{2}\)
4 \(a_{x}=-2 x\)
Explanation:
For motion to be S.H.M. acceleration of the particle must be proportional to negative of displacement. i.e., \(a \propto-(y\) or \(x)\) We should be clear that \(y\) has to be linear. Correct option is (4).
PHXI14:OSCILLATIONS
364295
The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is \(10\;{s^{ - 1}}\). At, \(t = 0\) the displacement is \(5\;m\). What is the maximum acceleration? The initial phase is \(\dfrac{\pi}{4}\).
1 \(500\;m/{s^2}\)
2 \(750\sqrt 2 \;m/{s^2}\)
3 \(750\;m/{s^2}\)
4 \(500\sqrt 2 \;m/{s^2}\)
Explanation:
Let \(a_{o}\) and \(v_{o}\) are maximum acceleration and velocity respectively. Given that \(\dfrac{a_{o}}{v_{o}}=10=\dfrac{\omega^{2} A}{\omega A}=\omega\) \(\omega=10 s^{-1}, \phi=\dfrac{\pi}{4} \& x=5 m\) Let \(\quad x=A \sin \left(\omega t+\dfrac{\pi}{4}\right)\) \(\begin{aligned}& 5=A \sin \left(0+\dfrac{\pi}{4}\right) \Rightarrow A=5 \\& a_{o}=\omega^{2} A=500 \sqrt{2}\end{aligned}\)
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PHXI14:OSCILLATIONS
364292
A particle moves such that its acceleration \(a\) is given by \(a=-b x\), where \(x\) is the displacement from equilibrium position and \(b\) is constant. The period of oscillation is
1 \(\dfrac{2 \pi}{\sqrt{b}}\)
2 \(2 \pi \sqrt{b}\)
3 \(2 \sqrt{\dfrac{\pi}{b}}\)
4 \(\dfrac{2 \pi}{b}\)
Explanation:
Given that \(a=-b x\) Comparing \(a=-\omega^{2} x \Rightarrow \omega^{2}=b \Rightarrow \omega=\sqrt{b}\) \( \Rightarrow \frac{{2\pi }}{T} = \sqrt b \Rightarrow T = \frac{{2\pi }}{{\sqrt b }}\)
PHXI14:OSCILLATIONS
364293
Assertion : In simple harmonic motion, the motion is to and fro and periodic. Reason : Velocity of the particle \((v)=\omega \sqrt{A^{2}-x^{2}}\) (where \(x\) is the displacement and \(A\) is amplitude)
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
A particle is said to execute simple harmonic motion, if it moves to and fro about a fixed point under the action of a restoring force which is directly proportional to its displacement from the fixed point and is always directed towards the fixed point. \(x=A \cos (\omega t+\phi)\) \(v=-A \omega \sin (\omega t+\phi)\) \(\Rightarrow v=\omega \sqrt{A^{2}-x^{2}}\) So, correct option is (2).
PHXI14:OSCILLATIONS
364294
The relation between acceleration and displacement of four particles are given below: Which, one of the particle is executing simple harmonic motion?
1 \(a_{x}=+2 x\)
2 \(a_{x}=+2 x^{2}\)
3 \(a_{x}=-2 x^{2}\)
4 \(a_{x}=-2 x\)
Explanation:
For motion to be S.H.M. acceleration of the particle must be proportional to negative of displacement. i.e., \(a \propto-(y\) or \(x)\) We should be clear that \(y\) has to be linear. Correct option is (4).
PHXI14:OSCILLATIONS
364295
The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is \(10\;{s^{ - 1}}\). At, \(t = 0\) the displacement is \(5\;m\). What is the maximum acceleration? The initial phase is \(\dfrac{\pi}{4}\).
1 \(500\;m/{s^2}\)
2 \(750\sqrt 2 \;m/{s^2}\)
3 \(750\;m/{s^2}\)
4 \(500\sqrt 2 \;m/{s^2}\)
Explanation:
Let \(a_{o}\) and \(v_{o}\) are maximum acceleration and velocity respectively. Given that \(\dfrac{a_{o}}{v_{o}}=10=\dfrac{\omega^{2} A}{\omega A}=\omega\) \(\omega=10 s^{-1}, \phi=\dfrac{\pi}{4} \& x=5 m\) Let \(\quad x=A \sin \left(\omega t+\dfrac{\pi}{4}\right)\) \(\begin{aligned}& 5=A \sin \left(0+\dfrac{\pi}{4}\right) \Rightarrow A=5 \\& a_{o}=\omega^{2} A=500 \sqrt{2}\end{aligned}\)
364292
A particle moves such that its acceleration \(a\) is given by \(a=-b x\), where \(x\) is the displacement from equilibrium position and \(b\) is constant. The period of oscillation is
1 \(\dfrac{2 \pi}{\sqrt{b}}\)
2 \(2 \pi \sqrt{b}\)
3 \(2 \sqrt{\dfrac{\pi}{b}}\)
4 \(\dfrac{2 \pi}{b}\)
Explanation:
Given that \(a=-b x\) Comparing \(a=-\omega^{2} x \Rightarrow \omega^{2}=b \Rightarrow \omega=\sqrt{b}\) \( \Rightarrow \frac{{2\pi }}{T} = \sqrt b \Rightarrow T = \frac{{2\pi }}{{\sqrt b }}\)
PHXI14:OSCILLATIONS
364293
Assertion : In simple harmonic motion, the motion is to and fro and periodic. Reason : Velocity of the particle \((v)=\omega \sqrt{A^{2}-x^{2}}\) (where \(x\) is the displacement and \(A\) is amplitude)
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
A particle is said to execute simple harmonic motion, if it moves to and fro about a fixed point under the action of a restoring force which is directly proportional to its displacement from the fixed point and is always directed towards the fixed point. \(x=A \cos (\omega t+\phi)\) \(v=-A \omega \sin (\omega t+\phi)\) \(\Rightarrow v=\omega \sqrt{A^{2}-x^{2}}\) So, correct option is (2).
PHXI14:OSCILLATIONS
364294
The relation between acceleration and displacement of four particles are given below: Which, one of the particle is executing simple harmonic motion?
1 \(a_{x}=+2 x\)
2 \(a_{x}=+2 x^{2}\)
3 \(a_{x}=-2 x^{2}\)
4 \(a_{x}=-2 x\)
Explanation:
For motion to be S.H.M. acceleration of the particle must be proportional to negative of displacement. i.e., \(a \propto-(y\) or \(x)\) We should be clear that \(y\) has to be linear. Correct option is (4).
PHXI14:OSCILLATIONS
364295
The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is \(10\;{s^{ - 1}}\). At, \(t = 0\) the displacement is \(5\;m\). What is the maximum acceleration? The initial phase is \(\dfrac{\pi}{4}\).
1 \(500\;m/{s^2}\)
2 \(750\sqrt 2 \;m/{s^2}\)
3 \(750\;m/{s^2}\)
4 \(500\sqrt 2 \;m/{s^2}\)
Explanation:
Let \(a_{o}\) and \(v_{o}\) are maximum acceleration and velocity respectively. Given that \(\dfrac{a_{o}}{v_{o}}=10=\dfrac{\omega^{2} A}{\omega A}=\omega\) \(\omega=10 s^{-1}, \phi=\dfrac{\pi}{4} \& x=5 m\) Let \(\quad x=A \sin \left(\omega t+\dfrac{\pi}{4}\right)\) \(\begin{aligned}& 5=A \sin \left(0+\dfrac{\pi}{4}\right) \Rightarrow A=5 \\& a_{o}=\omega^{2} A=500 \sqrt{2}\end{aligned}\)