364270
The average acceleration of a particle performing SHM over one complete oscillation is :
1 \(\dfrac{\omega^{2} A}{\sqrt{2}}\)
2 \(\dfrac{\omega^{2} A}{2}\)
3 \(A \omega^{2}\)
4 Zero
Explanation:
The average acceleration of particle performing SHM over one complete oscillation is zero. Because acceleration is a sinusoidal function and average value of sinusoidal function for complete cycle is zero.
PHXI14:OSCILLATIONS
364271
A body is in simple harmonic motion with time period half second \((T = 0.5\;s)\) and amplitude one \(cm(A = 1\;cm)\). Find the average velocity in the interval in which it moves from equilibrium position to half of its amplitude
1 \(6\;cm/s\)
2 \(4\;cm/s\)
3 \(12\;cm/s\)
4 \(16\;cm/s\)
Explanation:
Assume that the body starts its motion from the mean position. The displacement of the body is \(\begin{gathered}x=A \sin (\omega t) \\x=\sin \left(\dfrac{2 \pi}{0.5} t\right)=\sin (4 \pi t) \\v=\dfrac{d x}{d t}=4 \pi \cos (4 \pi t)\end{gathered}\) When its displacement is half \(\begin{aligned}& x=\dfrac{1}{2}=\sin \left(\dfrac{2 \pi}{0.5} t\right) \\& 4 \pi t=\dfrac{\pi}{6} \Rightarrow t=\dfrac{1}{24} s\end{aligned}\) The average velocity is \( < v > = \frac{{\int_0^t v dt}}{{t - 0}} = \frac{{4\pi \int_0^{\frac{1}{{24}}} {\cos } (4\pi t)dt}}{{\frac{1}{{24}}}}\) \( < v > = \frac{{96\pi }}{{4\pi }}\left[ {\sin \left( {4\pi r\left( {\frac{1}{{24}}} \right)} \right) - \sin (0)} \right]\) \( < v > = 12\;cm/s\)
JEE - 2014
PHXI14:OSCILLATIONS
364272
A body executing \(\mathrm{SHM}\) has velocity \(10\;cm{\rm{/}}s\) and \(7\;cm{\rm{/}}s\), when its displacements from the mean position are \(3\;cm\) and \(4\;cm\) respectively length of path
1 \(10\;cm\)
2 \(4.8\;cm\)
3 \(4\;cm\)
4 \(11.36\;cm\)
Explanation:
\(\dfrac{V_{1}}{V_{2}}=\dfrac{\omega \sqrt{A^{2}-Y_{1}^{2}}}{\omega \sqrt{A^{2}-Y_{2}^{2}}}\) \(\dfrac{10}{7}=\dfrac{\sqrt{A^{2}-9}}{\sqrt{A^{2}-1}}\) we get \(A = 4.76\;cm\).
PHXI14:OSCILLATIONS
364273
A particle executes \(\mathrm{SHM}\), its time period is \(16\,\,\sec \). If it passes through the centre of oscillation, then its velocity is \(2\;m{\rm{/}}s\) at time \(2\,\,\sec \). The amplitude will be:
364274
The displacement of particle varies with time as \(x = 12\sin \omega t - 16{\sin ^3}\omega t\,(in{\rm{ }}cm)\). If its motion is S.H.M., then its maximum acceleration is
364270
The average acceleration of a particle performing SHM over one complete oscillation is :
1 \(\dfrac{\omega^{2} A}{\sqrt{2}}\)
2 \(\dfrac{\omega^{2} A}{2}\)
3 \(A \omega^{2}\)
4 Zero
Explanation:
The average acceleration of particle performing SHM over one complete oscillation is zero. Because acceleration is a sinusoidal function and average value of sinusoidal function for complete cycle is zero.
PHXI14:OSCILLATIONS
364271
A body is in simple harmonic motion with time period half second \((T = 0.5\;s)\) and amplitude one \(cm(A = 1\;cm)\). Find the average velocity in the interval in which it moves from equilibrium position to half of its amplitude
1 \(6\;cm/s\)
2 \(4\;cm/s\)
3 \(12\;cm/s\)
4 \(16\;cm/s\)
Explanation:
Assume that the body starts its motion from the mean position. The displacement of the body is \(\begin{gathered}x=A \sin (\omega t) \\x=\sin \left(\dfrac{2 \pi}{0.5} t\right)=\sin (4 \pi t) \\v=\dfrac{d x}{d t}=4 \pi \cos (4 \pi t)\end{gathered}\) When its displacement is half \(\begin{aligned}& x=\dfrac{1}{2}=\sin \left(\dfrac{2 \pi}{0.5} t\right) \\& 4 \pi t=\dfrac{\pi}{6} \Rightarrow t=\dfrac{1}{24} s\end{aligned}\) The average velocity is \( < v > = \frac{{\int_0^t v dt}}{{t - 0}} = \frac{{4\pi \int_0^{\frac{1}{{24}}} {\cos } (4\pi t)dt}}{{\frac{1}{{24}}}}\) \( < v > = \frac{{96\pi }}{{4\pi }}\left[ {\sin \left( {4\pi r\left( {\frac{1}{{24}}} \right)} \right) - \sin (0)} \right]\) \( < v > = 12\;cm/s\)
JEE - 2014
PHXI14:OSCILLATIONS
364272
A body executing \(\mathrm{SHM}\) has velocity \(10\;cm{\rm{/}}s\) and \(7\;cm{\rm{/}}s\), when its displacements from the mean position are \(3\;cm\) and \(4\;cm\) respectively length of path
1 \(10\;cm\)
2 \(4.8\;cm\)
3 \(4\;cm\)
4 \(11.36\;cm\)
Explanation:
\(\dfrac{V_{1}}{V_{2}}=\dfrac{\omega \sqrt{A^{2}-Y_{1}^{2}}}{\omega \sqrt{A^{2}-Y_{2}^{2}}}\) \(\dfrac{10}{7}=\dfrac{\sqrt{A^{2}-9}}{\sqrt{A^{2}-1}}\) we get \(A = 4.76\;cm\).
PHXI14:OSCILLATIONS
364273
A particle executes \(\mathrm{SHM}\), its time period is \(16\,\,\sec \). If it passes through the centre of oscillation, then its velocity is \(2\;m{\rm{/}}s\) at time \(2\,\,\sec \). The amplitude will be:
364274
The displacement of particle varies with time as \(x = 12\sin \omega t - 16{\sin ^3}\omega t\,(in{\rm{ }}cm)\). If its motion is S.H.M., then its maximum acceleration is
364270
The average acceleration of a particle performing SHM over one complete oscillation is :
1 \(\dfrac{\omega^{2} A}{\sqrt{2}}\)
2 \(\dfrac{\omega^{2} A}{2}\)
3 \(A \omega^{2}\)
4 Zero
Explanation:
The average acceleration of particle performing SHM over one complete oscillation is zero. Because acceleration is a sinusoidal function and average value of sinusoidal function for complete cycle is zero.
PHXI14:OSCILLATIONS
364271
A body is in simple harmonic motion with time period half second \((T = 0.5\;s)\) and amplitude one \(cm(A = 1\;cm)\). Find the average velocity in the interval in which it moves from equilibrium position to half of its amplitude
1 \(6\;cm/s\)
2 \(4\;cm/s\)
3 \(12\;cm/s\)
4 \(16\;cm/s\)
Explanation:
Assume that the body starts its motion from the mean position. The displacement of the body is \(\begin{gathered}x=A \sin (\omega t) \\x=\sin \left(\dfrac{2 \pi}{0.5} t\right)=\sin (4 \pi t) \\v=\dfrac{d x}{d t}=4 \pi \cos (4 \pi t)\end{gathered}\) When its displacement is half \(\begin{aligned}& x=\dfrac{1}{2}=\sin \left(\dfrac{2 \pi}{0.5} t\right) \\& 4 \pi t=\dfrac{\pi}{6} \Rightarrow t=\dfrac{1}{24} s\end{aligned}\) The average velocity is \( < v > = \frac{{\int_0^t v dt}}{{t - 0}} = \frac{{4\pi \int_0^{\frac{1}{{24}}} {\cos } (4\pi t)dt}}{{\frac{1}{{24}}}}\) \( < v > = \frac{{96\pi }}{{4\pi }}\left[ {\sin \left( {4\pi r\left( {\frac{1}{{24}}} \right)} \right) - \sin (0)} \right]\) \( < v > = 12\;cm/s\)
JEE - 2014
PHXI14:OSCILLATIONS
364272
A body executing \(\mathrm{SHM}\) has velocity \(10\;cm{\rm{/}}s\) and \(7\;cm{\rm{/}}s\), when its displacements from the mean position are \(3\;cm\) and \(4\;cm\) respectively length of path
1 \(10\;cm\)
2 \(4.8\;cm\)
3 \(4\;cm\)
4 \(11.36\;cm\)
Explanation:
\(\dfrac{V_{1}}{V_{2}}=\dfrac{\omega \sqrt{A^{2}-Y_{1}^{2}}}{\omega \sqrt{A^{2}-Y_{2}^{2}}}\) \(\dfrac{10}{7}=\dfrac{\sqrt{A^{2}-9}}{\sqrt{A^{2}-1}}\) we get \(A = 4.76\;cm\).
PHXI14:OSCILLATIONS
364273
A particle executes \(\mathrm{SHM}\), its time period is \(16\,\,\sec \). If it passes through the centre of oscillation, then its velocity is \(2\;m{\rm{/}}s\) at time \(2\,\,\sec \). The amplitude will be:
364274
The displacement of particle varies with time as \(x = 12\sin \omega t - 16{\sin ^3}\omega t\,(in{\rm{ }}cm)\). If its motion is S.H.M., then its maximum acceleration is
364270
The average acceleration of a particle performing SHM over one complete oscillation is :
1 \(\dfrac{\omega^{2} A}{\sqrt{2}}\)
2 \(\dfrac{\omega^{2} A}{2}\)
3 \(A \omega^{2}\)
4 Zero
Explanation:
The average acceleration of particle performing SHM over one complete oscillation is zero. Because acceleration is a sinusoidal function and average value of sinusoidal function for complete cycle is zero.
PHXI14:OSCILLATIONS
364271
A body is in simple harmonic motion with time period half second \((T = 0.5\;s)\) and amplitude one \(cm(A = 1\;cm)\). Find the average velocity in the interval in which it moves from equilibrium position to half of its amplitude
1 \(6\;cm/s\)
2 \(4\;cm/s\)
3 \(12\;cm/s\)
4 \(16\;cm/s\)
Explanation:
Assume that the body starts its motion from the mean position. The displacement of the body is \(\begin{gathered}x=A \sin (\omega t) \\x=\sin \left(\dfrac{2 \pi}{0.5} t\right)=\sin (4 \pi t) \\v=\dfrac{d x}{d t}=4 \pi \cos (4 \pi t)\end{gathered}\) When its displacement is half \(\begin{aligned}& x=\dfrac{1}{2}=\sin \left(\dfrac{2 \pi}{0.5} t\right) \\& 4 \pi t=\dfrac{\pi}{6} \Rightarrow t=\dfrac{1}{24} s\end{aligned}\) The average velocity is \( < v > = \frac{{\int_0^t v dt}}{{t - 0}} = \frac{{4\pi \int_0^{\frac{1}{{24}}} {\cos } (4\pi t)dt}}{{\frac{1}{{24}}}}\) \( < v > = \frac{{96\pi }}{{4\pi }}\left[ {\sin \left( {4\pi r\left( {\frac{1}{{24}}} \right)} \right) - \sin (0)} \right]\) \( < v > = 12\;cm/s\)
JEE - 2014
PHXI14:OSCILLATIONS
364272
A body executing \(\mathrm{SHM}\) has velocity \(10\;cm{\rm{/}}s\) and \(7\;cm{\rm{/}}s\), when its displacements from the mean position are \(3\;cm\) and \(4\;cm\) respectively length of path
1 \(10\;cm\)
2 \(4.8\;cm\)
3 \(4\;cm\)
4 \(11.36\;cm\)
Explanation:
\(\dfrac{V_{1}}{V_{2}}=\dfrac{\omega \sqrt{A^{2}-Y_{1}^{2}}}{\omega \sqrt{A^{2}-Y_{2}^{2}}}\) \(\dfrac{10}{7}=\dfrac{\sqrt{A^{2}-9}}{\sqrt{A^{2}-1}}\) we get \(A = 4.76\;cm\).
PHXI14:OSCILLATIONS
364273
A particle executes \(\mathrm{SHM}\), its time period is \(16\,\,\sec \). If it passes through the centre of oscillation, then its velocity is \(2\;m{\rm{/}}s\) at time \(2\,\,\sec \). The amplitude will be:
364274
The displacement of particle varies with time as \(x = 12\sin \omega t - 16{\sin ^3}\omega t\,(in{\rm{ }}cm)\). If its motion is S.H.M., then its maximum acceleration is
364270
The average acceleration of a particle performing SHM over one complete oscillation is :
1 \(\dfrac{\omega^{2} A}{\sqrt{2}}\)
2 \(\dfrac{\omega^{2} A}{2}\)
3 \(A \omega^{2}\)
4 Zero
Explanation:
The average acceleration of particle performing SHM over one complete oscillation is zero. Because acceleration is a sinusoidal function and average value of sinusoidal function for complete cycle is zero.
PHXI14:OSCILLATIONS
364271
A body is in simple harmonic motion with time period half second \((T = 0.5\;s)\) and amplitude one \(cm(A = 1\;cm)\). Find the average velocity in the interval in which it moves from equilibrium position to half of its amplitude
1 \(6\;cm/s\)
2 \(4\;cm/s\)
3 \(12\;cm/s\)
4 \(16\;cm/s\)
Explanation:
Assume that the body starts its motion from the mean position. The displacement of the body is \(\begin{gathered}x=A \sin (\omega t) \\x=\sin \left(\dfrac{2 \pi}{0.5} t\right)=\sin (4 \pi t) \\v=\dfrac{d x}{d t}=4 \pi \cos (4 \pi t)\end{gathered}\) When its displacement is half \(\begin{aligned}& x=\dfrac{1}{2}=\sin \left(\dfrac{2 \pi}{0.5} t\right) \\& 4 \pi t=\dfrac{\pi}{6} \Rightarrow t=\dfrac{1}{24} s\end{aligned}\) The average velocity is \( < v > = \frac{{\int_0^t v dt}}{{t - 0}} = \frac{{4\pi \int_0^{\frac{1}{{24}}} {\cos } (4\pi t)dt}}{{\frac{1}{{24}}}}\) \( < v > = \frac{{96\pi }}{{4\pi }}\left[ {\sin \left( {4\pi r\left( {\frac{1}{{24}}} \right)} \right) - \sin (0)} \right]\) \( < v > = 12\;cm/s\)
JEE - 2014
PHXI14:OSCILLATIONS
364272
A body executing \(\mathrm{SHM}\) has velocity \(10\;cm{\rm{/}}s\) and \(7\;cm{\rm{/}}s\), when its displacements from the mean position are \(3\;cm\) and \(4\;cm\) respectively length of path
1 \(10\;cm\)
2 \(4.8\;cm\)
3 \(4\;cm\)
4 \(11.36\;cm\)
Explanation:
\(\dfrac{V_{1}}{V_{2}}=\dfrac{\omega \sqrt{A^{2}-Y_{1}^{2}}}{\omega \sqrt{A^{2}-Y_{2}^{2}}}\) \(\dfrac{10}{7}=\dfrac{\sqrt{A^{2}-9}}{\sqrt{A^{2}-1}}\) we get \(A = 4.76\;cm\).
PHXI14:OSCILLATIONS
364273
A particle executes \(\mathrm{SHM}\), its time period is \(16\,\,\sec \). If it passes through the centre of oscillation, then its velocity is \(2\;m{\rm{/}}s\) at time \(2\,\,\sec \). The amplitude will be:
364274
The displacement of particle varies with time as \(x = 12\sin \omega t - 16{\sin ^3}\omega t\,(in{\rm{ }}cm)\). If its motion is S.H.M., then its maximum acceleration is
