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PHXI14:OSCILLATIONS

364330 A piston is performing S.H.M. in the vertical direction with a frequency of \(0.5\;Hz\). A block of \(10\;kg\) is placed on the piston. The maximum amplitude of the system such that the block remains in contact with the piston is

1 \(1.5\;m\)

2 \(1\;m\)

3 \(0.1\;m\)

4 \(0.5\;m\)

KCET - 2019

PHXI14:OSCILLATIONS

364331 Find the period of small oscillations in a horizontal plane performed by a ball of mass \(m=\dfrac{4}{10 \pi^{2}} {~kg}\), fixed at the middle of a horizontally stretched string \(l=1.0 {~m}\) in length. The tension of the string is assumed to be constant and equal to \(T=10 {~N}\).

1 \(0.2\,s\)

2 \(0.8\,s\)

3 \(0.10\,s\)

4 \(0.5\,s\)

PHXI14:OSCILLATIONS

364332 A sphere of mass \(m\) and radius \(r\) rolls without slipping on a rough concave surface of large radius \(\mathrm{R}\). It makes small oscillations about the lowest point. Find the time period.

1 \(2 \pi \sqrt{\dfrac{(R-r)}{g}}\)

2 \(2 \pi \sqrt{\dfrac{5(R-r)}{2 g}}\)

3 \(2 \pi \sqrt{\dfrac{2(R-r)}{5 g}}\)

4 \(2 \pi \sqrt{\dfrac{7(R-r)}{5 g}}\)

Explanation:

Let at any instant, the body is at angular position \(\theta\) with respect to the vertical line drawn from the centre \(Q\). If \(\phi\) is the angular displacement of the ball about its centre, then
\((R-r) \theta=r \phi \quad \therefore \theta=\left(\dfrac{r}{R-r}\right) \phi\)
supporting img
(b)Restoring torque acting on the ball
\(\tau=-m g \sin \theta \times r\)
For small \(\theta, \quad \sin \theta=\theta\)
\(\begin{gathered}\therefore \tau=-m g \theta \times r \\\text { or } \tau=-m g\left[\dfrac{r}{R-r}\right] \phi r=m g\left(\dfrac{r^{2}}{R-r}\right)(-\phi)\end{gathered}\)
Angular acceleration \(\alpha=\dfrac{m g}{I}\left(\dfrac{r^{2}}{R-r}\right)(-\phi)\)
Now comparing above equation with standard equation of SHM,
\(\alpha=-\omega^{2} \phi, \text { we get } \omega=\sqrt{\dfrac{m g}{I}\left(\dfrac{r^{2}}{R-r}\right)}\)
Here \(I\) is the moment of inertia of the rolling ball about the point of contact which is \(I=(7 / 5) m r^{2}\)
\(\omega=\sqrt{\dfrac{m g}{\dfrac{7}{5} m r^{2}}\left(\dfrac{r^{2}}{R-r}\right)}=\sqrt{\dfrac{5}{7} \dfrac{g}{(R-r)}}\)
and \(\quad T=\dfrac{2 \pi}{\omega}=2 \pi \sqrt{\dfrac{7(R-r)}{5 g}}\)
Alternative method:
We can imagine that when the \(CM\) of the body is shifted slightly by an angle \(\theta\) relative to the centre of curvature of the surface, the magnitude of acceleration of the \(CM\) of the body can be given as
\(a=\dfrac{g \sin \theta}{1+\dfrac{k^{2}}{r^{2}}}\)
(because the body is accelerating down the inclined plane of instantaneous angle \(\theta\) )
Substituting \(\sin \theta=\theta=\dfrac{x}{R-r}\), we have
\(a = \frac{{ - gx}}{{(R - r)\left( {1 + \frac{{{k^2}}}{{{r^2}}}} \right)}}\)
supporting img
(negative sign is given because \(\vec{a}\) and \(\vec{x}\) are oppositely directed)
Comparing the above equation with \(a=-\omega^{2} x\), we have
\(\omega^{2}=\dfrac{g}{(R-r)\left(1+\dfrac{k^{2}}{r^{2}}\right)}\)
For solid sphere \(k^{2}=\dfrac{2 r^{2}}{5}\).
Hence \(\omega=\sqrt{\dfrac{5}{7} \dfrac{g}{(R-r)}}\)

PHXI14:OSCILLATIONS

364333 Assertion :
If a hole were drilled through the centre of earth and ball dropped into the hole at one end, it will oscillate.
Reason :
The ball will come out of other end if passed through centre hole in earth.

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.

PHXI14:OSCILLATIONS

364330 A piston is performing S.H.M. in the vertical direction with a frequency of \(0.5\;Hz\). A block of \(10\;kg\) is placed on the piston. The maximum amplitude of the system such that the block remains in contact with the piston is

1 \(1.5\;m\)

2 \(1\;m\)

3 \(0.1\;m\)

4 \(0.5\;m\)

KCET - 2019

PHXI14:OSCILLATIONS

364331 Find the period of small oscillations in a horizontal plane performed by a ball of mass \(m=\dfrac{4}{10 \pi^{2}} {~kg}\), fixed at the middle of a horizontally stretched string \(l=1.0 {~m}\) in length. The tension of the string is assumed to be constant and equal to \(T=10 {~N}\).

1 \(0.2\,s\)

2 \(0.8\,s\)

3 \(0.10\,s\)

4 \(0.5\,s\)

PHXI14:OSCILLATIONS

364332 A sphere of mass \(m\) and radius \(r\) rolls without slipping on a rough concave surface of large radius \(\mathrm{R}\). It makes small oscillations about the lowest point. Find the time period.

1 \(2 \pi \sqrt{\dfrac{(R-r)}{g}}\)

2 \(2 \pi \sqrt{\dfrac{5(R-r)}{2 g}}\)

3 \(2 \pi \sqrt{\dfrac{2(R-r)}{5 g}}\)

4 \(2 \pi \sqrt{\dfrac{7(R-r)}{5 g}}\)

Explanation:

Let at any instant, the body is at angular position \(\theta\) with respect to the vertical line drawn from the centre \(Q\). If \(\phi\) is the angular displacement of the ball about its centre, then
\((R-r) \theta=r \phi \quad \therefore \theta=\left(\dfrac{r}{R-r}\right) \phi\)
supporting img
(b)Restoring torque acting on the ball
\(\tau=-m g \sin \theta \times r\)
For small \(\theta, \quad \sin \theta=\theta\)
\(\begin{gathered}\therefore \tau=-m g \theta \times r \\\text { or } \tau=-m g\left[\dfrac{r}{R-r}\right] \phi r=m g\left(\dfrac{r^{2}}{R-r}\right)(-\phi)\end{gathered}\)
Angular acceleration \(\alpha=\dfrac{m g}{I}\left(\dfrac{r^{2}}{R-r}\right)(-\phi)\)
Now comparing above equation with standard equation of SHM,
\(\alpha=-\omega^{2} \phi, \text { we get } \omega=\sqrt{\dfrac{m g}{I}\left(\dfrac{r^{2}}{R-r}\right)}\)
Here \(I\) is the moment of inertia of the rolling ball about the point of contact which is \(I=(7 / 5) m r^{2}\)
\(\omega=\sqrt{\dfrac{m g}{\dfrac{7}{5} m r^{2}}\left(\dfrac{r^{2}}{R-r}\right)}=\sqrt{\dfrac{5}{7} \dfrac{g}{(R-r)}}\)
and \(\quad T=\dfrac{2 \pi}{\omega}=2 \pi \sqrt{\dfrac{7(R-r)}{5 g}}\)
Alternative method:
We can imagine that when the \(CM\) of the body is shifted slightly by an angle \(\theta\) relative to the centre of curvature of the surface, the magnitude of acceleration of the \(CM\) of the body can be given as
\(a=\dfrac{g \sin \theta}{1+\dfrac{k^{2}}{r^{2}}}\)
(because the body is accelerating down the inclined plane of instantaneous angle \(\theta\) )
Substituting \(\sin \theta=\theta=\dfrac{x}{R-r}\), we have
\(a = \frac{{ - gx}}{{(R - r)\left( {1 + \frac{{{k^2}}}{{{r^2}}}} \right)}}\)
supporting img
(negative sign is given because \(\vec{a}\) and \(\vec{x}\) are oppositely directed)
Comparing the above equation with \(a=-\omega^{2} x\), we have
\(\omega^{2}=\dfrac{g}{(R-r)\left(1+\dfrac{k^{2}}{r^{2}}\right)}\)
For solid sphere \(k^{2}=\dfrac{2 r^{2}}{5}\).
Hence \(\omega=\sqrt{\dfrac{5}{7} \dfrac{g}{(R-r)}}\)

PHXI14:OSCILLATIONS

364333 Assertion :
If a hole were drilled through the centre of earth and ball dropped into the hole at one end, it will oscillate.
Reason :
The ball will come out of other end if passed through centre hole in earth.

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.

PHXI14:OSCILLATIONS

364330 A piston is performing S.H.M. in the vertical direction with a frequency of \(0.5\;Hz\). A block of \(10\;kg\) is placed on the piston. The maximum amplitude of the system such that the block remains in contact with the piston is

1 \(1.5\;m\)

2 \(1\;m\)

3 \(0.1\;m\)

4 \(0.5\;m\)

KCET - 2019

PHXI14:OSCILLATIONS

364331 Find the period of small oscillations in a horizontal plane performed by a ball of mass \(m=\dfrac{4}{10 \pi^{2}} {~kg}\), fixed at the middle of a horizontally stretched string \(l=1.0 {~m}\) in length. The tension of the string is assumed to be constant and equal to \(T=10 {~N}\).

1 \(0.2\,s\)

2 \(0.8\,s\)

3 \(0.10\,s\)

4 \(0.5\,s\)

PHXI14:OSCILLATIONS

364332 A sphere of mass \(m\) and radius \(r\) rolls without slipping on a rough concave surface of large radius \(\mathrm{R}\). It makes small oscillations about the lowest point. Find the time period.

1 \(2 \pi \sqrt{\dfrac{(R-r)}{g}}\)

2 \(2 \pi \sqrt{\dfrac{5(R-r)}{2 g}}\)

3 \(2 \pi \sqrt{\dfrac{2(R-r)}{5 g}}\)

4 \(2 \pi \sqrt{\dfrac{7(R-r)}{5 g}}\)

Explanation:

Let at any instant, the body is at angular position \(\theta\) with respect to the vertical line drawn from the centre \(Q\). If \(\phi\) is the angular displacement of the ball about its centre, then
\((R-r) \theta=r \phi \quad \therefore \theta=\left(\dfrac{r}{R-r}\right) \phi\)
supporting img
(b)Restoring torque acting on the ball
\(\tau=-m g \sin \theta \times r\)
For small \(\theta, \quad \sin \theta=\theta\)
\(\begin{gathered}\therefore \tau=-m g \theta \times r \\\text { or } \tau=-m g\left[\dfrac{r}{R-r}\right] \phi r=m g\left(\dfrac{r^{2}}{R-r}\right)(-\phi)\end{gathered}\)
Angular acceleration \(\alpha=\dfrac{m g}{I}\left(\dfrac{r^{2}}{R-r}\right)(-\phi)\)
Now comparing above equation with standard equation of SHM,
\(\alpha=-\omega^{2} \phi, \text { we get } \omega=\sqrt{\dfrac{m g}{I}\left(\dfrac{r^{2}}{R-r}\right)}\)
Here \(I\) is the moment of inertia of the rolling ball about the point of contact which is \(I=(7 / 5) m r^{2}\)
\(\omega=\sqrt{\dfrac{m g}{\dfrac{7}{5} m r^{2}}\left(\dfrac{r^{2}}{R-r}\right)}=\sqrt{\dfrac{5}{7} \dfrac{g}{(R-r)}}\)
and \(\quad T=\dfrac{2 \pi}{\omega}=2 \pi \sqrt{\dfrac{7(R-r)}{5 g}}\)
Alternative method:
We can imagine that when the \(CM\) of the body is shifted slightly by an angle \(\theta\) relative to the centre of curvature of the surface, the magnitude of acceleration of the \(CM\) of the body can be given as
\(a=\dfrac{g \sin \theta}{1+\dfrac{k^{2}}{r^{2}}}\)
(because the body is accelerating down the inclined plane of instantaneous angle \(\theta\) )
Substituting \(\sin \theta=\theta=\dfrac{x}{R-r}\), we have
\(a = \frac{{ - gx}}{{(R - r)\left( {1 + \frac{{{k^2}}}{{{r^2}}}} \right)}}\)
supporting img
(negative sign is given because \(\vec{a}\) and \(\vec{x}\) are oppositely directed)
Comparing the above equation with \(a=-\omega^{2} x\), we have
\(\omega^{2}=\dfrac{g}{(R-r)\left(1+\dfrac{k^{2}}{r^{2}}\right)}\)
For solid sphere \(k^{2}=\dfrac{2 r^{2}}{5}\).
Hence \(\omega=\sqrt{\dfrac{5}{7} \dfrac{g}{(R-r)}}\)

PHXI14:OSCILLATIONS

364333 Assertion :
If a hole were drilled through the centre of earth and ball dropped into the hole at one end, it will oscillate.
Reason :
The ball will come out of other end if passed through centre hole in earth.

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.

PHXI14:OSCILLATIONS

364330 A piston is performing S.H.M. in the vertical direction with a frequency of \(0.5\;Hz\). A block of \(10\;kg\) is placed on the piston. The maximum amplitude of the system such that the block remains in contact with the piston is

1 \(1.5\;m\)

2 \(1\;m\)

3 \(0.1\;m\)

4 \(0.5\;m\)

KCET - 2019

PHXI14:OSCILLATIONS

364331 Find the period of small oscillations in a horizontal plane performed by a ball of mass \(m=\dfrac{4}{10 \pi^{2}} {~kg}\), fixed at the middle of a horizontally stretched string \(l=1.0 {~m}\) in length. The tension of the string is assumed to be constant and equal to \(T=10 {~N}\).

1 \(0.2\,s\)

2 \(0.8\,s\)

3 \(0.10\,s\)

4 \(0.5\,s\)

PHXI14:OSCILLATIONS

364332 A sphere of mass \(m\) and radius \(r\) rolls without slipping on a rough concave surface of large radius \(\mathrm{R}\). It makes small oscillations about the lowest point. Find the time period.

1 \(2 \pi \sqrt{\dfrac{(R-r)}{g}}\)

2 \(2 \pi \sqrt{\dfrac{5(R-r)}{2 g}}\)

3 \(2 \pi \sqrt{\dfrac{2(R-r)}{5 g}}\)

4 \(2 \pi \sqrt{\dfrac{7(R-r)}{5 g}}\)

Explanation:

Let at any instant, the body is at angular position \(\theta\) with respect to the vertical line drawn from the centre \(Q\). If \(\phi\) is the angular displacement of the ball about its centre, then
\((R-r) \theta=r \phi \quad \therefore \theta=\left(\dfrac{r}{R-r}\right) \phi\)
supporting img
(b)Restoring torque acting on the ball
\(\tau=-m g \sin \theta \times r\)
For small \(\theta, \quad \sin \theta=\theta\)
\(\begin{gathered}\therefore \tau=-m g \theta \times r \\\text { or } \tau=-m g\left[\dfrac{r}{R-r}\right] \phi r=m g\left(\dfrac{r^{2}}{R-r}\right)(-\phi)\end{gathered}\)
Angular acceleration \(\alpha=\dfrac{m g}{I}\left(\dfrac{r^{2}}{R-r}\right)(-\phi)\)
Now comparing above equation with standard equation of SHM,
\(\alpha=-\omega^{2} \phi, \text { we get } \omega=\sqrt{\dfrac{m g}{I}\left(\dfrac{r^{2}}{R-r}\right)}\)
Here \(I\) is the moment of inertia of the rolling ball about the point of contact which is \(I=(7 / 5) m r^{2}\)
\(\omega=\sqrt{\dfrac{m g}{\dfrac{7}{5} m r^{2}}\left(\dfrac{r^{2}}{R-r}\right)}=\sqrt{\dfrac{5}{7} \dfrac{g}{(R-r)}}\)
and \(\quad T=\dfrac{2 \pi}{\omega}=2 \pi \sqrt{\dfrac{7(R-r)}{5 g}}\)
Alternative method:
We can imagine that when the \(CM\) of the body is shifted slightly by an angle \(\theta\) relative to the centre of curvature of the surface, the magnitude of acceleration of the \(CM\) of the body can be given as
\(a=\dfrac{g \sin \theta}{1+\dfrac{k^{2}}{r^{2}}}\)
(because the body is accelerating down the inclined plane of instantaneous angle \(\theta\) )
Substituting \(\sin \theta=\theta=\dfrac{x}{R-r}\), we have
\(a = \frac{{ - gx}}{{(R - r)\left( {1 + \frac{{{k^2}}}{{{r^2}}}} \right)}}\)
supporting img
(negative sign is given because \(\vec{a}\) and \(\vec{x}\) are oppositely directed)
Comparing the above equation with \(a=-\omega^{2} x\), we have
\(\omega^{2}=\dfrac{g}{(R-r)\left(1+\dfrac{k^{2}}{r^{2}}\right)}\)
For solid sphere \(k^{2}=\dfrac{2 r^{2}}{5}\).
Hence \(\omega=\sqrt{\dfrac{5}{7} \dfrac{g}{(R-r)}}\)

PHXI14:OSCILLATIONS

364333 Assertion :
If a hole were drilled through the centre of earth and ball dropped into the hole at one end, it will oscillate.
Reason :
The ball will come out of other end if passed through centre hole in earth.

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.