364300
Which one of the following statements is true for the speed ' \(v\) ' and the acceleration ' \(a\) ' of a particle executing simple harmonic motion
1 When ' \(v\) ' is zero, \(a\) is zero
2 Value of \(a\) is zero, whatever may be the value of ' \(v\) '
3 When ' \(v\) ' is maximum, \(a\) is maximum
4 When ' \(v\) ' is maximum, \(a\) is zero
Explanation:
In SHM if \(v=v_{\text {max }}\) then \(a=0\) if \(v=0\) then \(a=a_{\max }\)
PHXI14:OSCILLATIONS
364301
Assertion : Simple harmonic motion is not a uniformly accelerated motion. Reason : Velocity is non-uniform in SHM.
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:
In simple harmonic motion (SHM), the acceleration (a) varies with position (y), indicating non-uniform acceleration. \(y=A \cos (\omega t+\phi)\) \(\Rightarrow v\) and \(a\) will be varying So correct option is (2).
PHXI14:OSCILLATIONS
364302
If \(x, v\) and a denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period \(T\), then, which of the following does not change with time?
1 \(a T / x\)
2 \(a T / 2 \pi v\)
3 \(a T / v\)
4 \(a^{2} T^{2} / 4 \pi^{2} v^{2}\)
Explanation:
For an SHM, the acceleration \(a=-\omega^{2} x\) where \(\omega^{2}\) is a constant. Therefore, \(\dfrac{a}{x}\) is a constant. The time period \(\mathrm{T}\) is also constant. Therefore, \(\dfrac{a T}{x}\) is a constant.
PHXI14:OSCILLATIONS
364303
In SHM there is always a constant ratio between the displacement of the body and its
1 Velocity
2 Acceleration
3 Mass of the particle
4 All of the above
Explanation:
Conceptual Question
PHXI14:OSCILLATIONS
364304
If velocity of a particle is given as \(v=\pi \sin \omega t\) then find the average value of velocity from 0 to \(T/2\) time interval. where \(\omega\) and \(T\) are angular velocity and time period respectively :-
1 \(\dfrac{\pi}{\sqrt{2}}\)
2 \(\dfrac{\pi}{2}\)
3 2
4 \(2 \sqrt{2}\)
Explanation:
\(v_{\text {avg }}=\dfrac{\int_{0}^{T / 2} \pi \sin \omega t d t}{\int_{0}^{T / 2} d t}=-\dfrac{2 \pi}{\omega T}[\cos \omega t]_{0}^{T / 2}=2\)
364300
Which one of the following statements is true for the speed ' \(v\) ' and the acceleration ' \(a\) ' of a particle executing simple harmonic motion
1 When ' \(v\) ' is zero, \(a\) is zero
2 Value of \(a\) is zero, whatever may be the value of ' \(v\) '
3 When ' \(v\) ' is maximum, \(a\) is maximum
4 When ' \(v\) ' is maximum, \(a\) is zero
Explanation:
In SHM if \(v=v_{\text {max }}\) then \(a=0\) if \(v=0\) then \(a=a_{\max }\)
PHXI14:OSCILLATIONS
364301
Assertion : Simple harmonic motion is not a uniformly accelerated motion. Reason : Velocity is non-uniform in SHM.
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:
In simple harmonic motion (SHM), the acceleration (a) varies with position (y), indicating non-uniform acceleration. \(y=A \cos (\omega t+\phi)\) \(\Rightarrow v\) and \(a\) will be varying So correct option is (2).
PHXI14:OSCILLATIONS
364302
If \(x, v\) and a denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period \(T\), then, which of the following does not change with time?
1 \(a T / x\)
2 \(a T / 2 \pi v\)
3 \(a T / v\)
4 \(a^{2} T^{2} / 4 \pi^{2} v^{2}\)
Explanation:
For an SHM, the acceleration \(a=-\omega^{2} x\) where \(\omega^{2}\) is a constant. Therefore, \(\dfrac{a}{x}\) is a constant. The time period \(\mathrm{T}\) is also constant. Therefore, \(\dfrac{a T}{x}\) is a constant.
PHXI14:OSCILLATIONS
364303
In SHM there is always a constant ratio between the displacement of the body and its
1 Velocity
2 Acceleration
3 Mass of the particle
4 All of the above
Explanation:
Conceptual Question
PHXI14:OSCILLATIONS
364304
If velocity of a particle is given as \(v=\pi \sin \omega t\) then find the average value of velocity from 0 to \(T/2\) time interval. where \(\omega\) and \(T\) are angular velocity and time period respectively :-
1 \(\dfrac{\pi}{\sqrt{2}}\)
2 \(\dfrac{\pi}{2}\)
3 2
4 \(2 \sqrt{2}\)
Explanation:
\(v_{\text {avg }}=\dfrac{\int_{0}^{T / 2} \pi \sin \omega t d t}{\int_{0}^{T / 2} d t}=-\dfrac{2 \pi}{\omega T}[\cos \omega t]_{0}^{T / 2}=2\)
364300
Which one of the following statements is true for the speed ' \(v\) ' and the acceleration ' \(a\) ' of a particle executing simple harmonic motion
1 When ' \(v\) ' is zero, \(a\) is zero
2 Value of \(a\) is zero, whatever may be the value of ' \(v\) '
3 When ' \(v\) ' is maximum, \(a\) is maximum
4 When ' \(v\) ' is maximum, \(a\) is zero
Explanation:
In SHM if \(v=v_{\text {max }}\) then \(a=0\) if \(v=0\) then \(a=a_{\max }\)
PHXI14:OSCILLATIONS
364301
Assertion : Simple harmonic motion is not a uniformly accelerated motion. Reason : Velocity is non-uniform in SHM.
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:
In simple harmonic motion (SHM), the acceleration (a) varies with position (y), indicating non-uniform acceleration. \(y=A \cos (\omega t+\phi)\) \(\Rightarrow v\) and \(a\) will be varying So correct option is (2).
PHXI14:OSCILLATIONS
364302
If \(x, v\) and a denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period \(T\), then, which of the following does not change with time?
1 \(a T / x\)
2 \(a T / 2 \pi v\)
3 \(a T / v\)
4 \(a^{2} T^{2} / 4 \pi^{2} v^{2}\)
Explanation:
For an SHM, the acceleration \(a=-\omega^{2} x\) where \(\omega^{2}\) is a constant. Therefore, \(\dfrac{a}{x}\) is a constant. The time period \(\mathrm{T}\) is also constant. Therefore, \(\dfrac{a T}{x}\) is a constant.
PHXI14:OSCILLATIONS
364303
In SHM there is always a constant ratio between the displacement of the body and its
1 Velocity
2 Acceleration
3 Mass of the particle
4 All of the above
Explanation:
Conceptual Question
PHXI14:OSCILLATIONS
364304
If velocity of a particle is given as \(v=\pi \sin \omega t\) then find the average value of velocity from 0 to \(T/2\) time interval. where \(\omega\) and \(T\) are angular velocity and time period respectively :-
1 \(\dfrac{\pi}{\sqrt{2}}\)
2 \(\dfrac{\pi}{2}\)
3 2
4 \(2 \sqrt{2}\)
Explanation:
\(v_{\text {avg }}=\dfrac{\int_{0}^{T / 2} \pi \sin \omega t d t}{\int_{0}^{T / 2} d t}=-\dfrac{2 \pi}{\omega T}[\cos \omega t]_{0}^{T / 2}=2\)
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PHXI14:OSCILLATIONS
364300
Which one of the following statements is true for the speed ' \(v\) ' and the acceleration ' \(a\) ' of a particle executing simple harmonic motion
1 When ' \(v\) ' is zero, \(a\) is zero
2 Value of \(a\) is zero, whatever may be the value of ' \(v\) '
3 When ' \(v\) ' is maximum, \(a\) is maximum
4 When ' \(v\) ' is maximum, \(a\) is zero
Explanation:
In SHM if \(v=v_{\text {max }}\) then \(a=0\) if \(v=0\) then \(a=a_{\max }\)
PHXI14:OSCILLATIONS
364301
Assertion : Simple harmonic motion is not a uniformly accelerated motion. Reason : Velocity is non-uniform in SHM.
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:
In simple harmonic motion (SHM), the acceleration (a) varies with position (y), indicating non-uniform acceleration. \(y=A \cos (\omega t+\phi)\) \(\Rightarrow v\) and \(a\) will be varying So correct option is (2).
PHXI14:OSCILLATIONS
364302
If \(x, v\) and a denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period \(T\), then, which of the following does not change with time?
1 \(a T / x\)
2 \(a T / 2 \pi v\)
3 \(a T / v\)
4 \(a^{2} T^{2} / 4 \pi^{2} v^{2}\)
Explanation:
For an SHM, the acceleration \(a=-\omega^{2} x\) where \(\omega^{2}\) is a constant. Therefore, \(\dfrac{a}{x}\) is a constant. The time period \(\mathrm{T}\) is also constant. Therefore, \(\dfrac{a T}{x}\) is a constant.
PHXI14:OSCILLATIONS
364303
In SHM there is always a constant ratio between the displacement of the body and its
1 Velocity
2 Acceleration
3 Mass of the particle
4 All of the above
Explanation:
Conceptual Question
PHXI14:OSCILLATIONS
364304
If velocity of a particle is given as \(v=\pi \sin \omega t\) then find the average value of velocity from 0 to \(T/2\) time interval. where \(\omega\) and \(T\) are angular velocity and time period respectively :-
1 \(\dfrac{\pi}{\sqrt{2}}\)
2 \(\dfrac{\pi}{2}\)
3 2
4 \(2 \sqrt{2}\)
Explanation:
\(v_{\text {avg }}=\dfrac{\int_{0}^{T / 2} \pi \sin \omega t d t}{\int_{0}^{T / 2} d t}=-\dfrac{2 \pi}{\omega T}[\cos \omega t]_{0}^{T / 2}=2\)
364300
Which one of the following statements is true for the speed ' \(v\) ' and the acceleration ' \(a\) ' of a particle executing simple harmonic motion
1 When ' \(v\) ' is zero, \(a\) is zero
2 Value of \(a\) is zero, whatever may be the value of ' \(v\) '
3 When ' \(v\) ' is maximum, \(a\) is maximum
4 When ' \(v\) ' is maximum, \(a\) is zero
Explanation:
In SHM if \(v=v_{\text {max }}\) then \(a=0\) if \(v=0\) then \(a=a_{\max }\)
PHXI14:OSCILLATIONS
364301
Assertion : Simple harmonic motion is not a uniformly accelerated motion. Reason : Velocity is non-uniform in SHM.
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:
In simple harmonic motion (SHM), the acceleration (a) varies with position (y), indicating non-uniform acceleration. \(y=A \cos (\omega t+\phi)\) \(\Rightarrow v\) and \(a\) will be varying So correct option is (2).
PHXI14:OSCILLATIONS
364302
If \(x, v\) and a denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period \(T\), then, which of the following does not change with time?
1 \(a T / x\)
2 \(a T / 2 \pi v\)
3 \(a T / v\)
4 \(a^{2} T^{2} / 4 \pi^{2} v^{2}\)
Explanation:
For an SHM, the acceleration \(a=-\omega^{2} x\) where \(\omega^{2}\) is a constant. Therefore, \(\dfrac{a}{x}\) is a constant. The time period \(\mathrm{T}\) is also constant. Therefore, \(\dfrac{a T}{x}\) is a constant.
PHXI14:OSCILLATIONS
364303
In SHM there is always a constant ratio between the displacement of the body and its
1 Velocity
2 Acceleration
3 Mass of the particle
4 All of the above
Explanation:
Conceptual Question
PHXI14:OSCILLATIONS
364304
If velocity of a particle is given as \(v=\pi \sin \omega t\) then find the average value of velocity from 0 to \(T/2\) time interval. where \(\omega\) and \(T\) are angular velocity and time period respectively :-
1 \(\dfrac{\pi}{\sqrt{2}}\)
2 \(\dfrac{\pi}{2}\)
3 2
4 \(2 \sqrt{2}\)
Explanation:
\(v_{\text {avg }}=\dfrac{\int_{0}^{T / 2} \pi \sin \omega t d t}{\int_{0}^{T / 2} d t}=-\dfrac{2 \pi}{\omega T}[\cos \omega t]_{0}^{T / 2}=2\)