364325
A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency \(\omega\). The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time
1 For an amplitude of \(g^{2} / \omega^{2}\)
2 For an amplitude of \(g / \omega^{2}\)
3 At the mean position of the platform
4 At the highest position of the platform
Explanation:
The coin will leave contact when it is at the highest point and for that condition Maximum acceleration \(=\) Acceleration due to gravity \(\Rightarrow \omega^{2} A=g \quad \Rightarrow\) Amplitude \(A=\dfrac{g}{\omega^{2}}\).
PHXI14:OSCILLATIONS
364326
Consider a liquid which fills a uniform U-tube, as shown in fig up to a height \(h\). Find angular frequency of small oscillations of the liquid in the U-tube.
1 \(\sqrt{g / 2 h}\)
2 \(\sqrt{g / h}\)
3 \(\sqrt{2 g / h}\)
4 \(\sqrt{3 g / 2 h}\)
Explanation:
Suppose that the liquid is displaced slightly from equilibrium so that its level rises in one arm of the tube, while it is depressed in the second arm by the same amount, \(x\). If the density of the liquid is \(\rho\), then the total mechanical energy of the liquid column is The difference in the two levels is \(2 x\). So the column having length \(2 x\) produces restoring force on the entire liquid. \(\begin{aligned}& \rho A 2 x g=\rho A 2 h a \\& a=-\left(\dfrac{g}{h}\right) x \Rightarrow \omega=\sqrt{\dfrac{g}{h}}\end{aligned}\)
PHXI14:OSCILLATIONS
364327
A clock \(S\) is based on oscillation of a spring and a clock \(P\) is based on pendulum motion. Both clocks run at the same rate on the earth. On a planet having the same density as earth but twice the radius
1 \(S\) will run faster than \(P\)
2 \(P\) will run faster than \(S\)
3 both will run at the same rate as on the earth
4 both will run at the same rate which will be different from that on the earth
Explanation:
Acceleration due to gravity, \(g=\dfrac{G M}{R^{2}}=\dfrac{G \times \dfrac{4}{3} \pi R^{3} \rho}{R^{2}}=\dfrac{4}{3} \pi G \rho R\) \(\Rightarrow g \propto R\) For pendulum clock, \(g\) will increase on the planet so time period will decrease. But for spring clock, it will not change. Hence, \(P\) will run faster than \(S\)
PHXI14:OSCILLATIONS
364328
A particle under the action of a SHM has a period of \(3\,{\rm{sec}}\) and under the effect of another it has a period \(4\,{\rm{sec}}\). What will be its period under the combined action of both the SHM's in the same direction?
1 \(7\,\sec \)
2 \(5\,\sec \)
3 \(2.4\,\sec \)
4 \(0.4\,\sec \)
Explanation:
\(T=2 \pi \sqrt{\dfrac{m}{K}}\) or \(T \propto \dfrac{1}{\sqrt{K}}\) \(T\) has increased \(\dfrac{4}{3}\) times. \(K^{\prime}=\dfrac{9 K}{16}\) When both are combined \(\begin{aligned}K_{n e t} & =K+K_{1} \\& =\dfrac{25}{16} K\end{aligned}\) New time period will become \(\dfrac{4}{3}\) times of \(3\,{\rm{sec}}\) or \(2.4\,{\rm{sec}}\)
PHXI14:OSCILLATIONS
364329
An air chamber of volume \(V\), has a long neck of cross-sectional area A. A ball of mass \(m\) is fitted smoothly in the neck. The bulk modulus of air is \(B\). If the ball is pressed down slightly and released, the time period of its oscillation is
1 \(\pi \sqrt{\dfrac{2 m V}{B A^{2}}}\)
2 \(2 \pi \sqrt{\dfrac{m V}{2 B A^{2}}}\)
3 \(\dfrac{\pi}{2} \sqrt{\dfrac{m}{B A^{2}}}\)
4 \(2 \pi \sqrt{\dfrac{m V}{B A^{2}}}\)
Explanation:
Let the ball is pressed down by a small amount \(\mathrm{y}\), then the volume of air decreases by Ay. Then, excess pressure is \(d p=-B\left(\dfrac{d V}{V}\right)=-B\left(\dfrac{y A}{V}\right)\) A restoring force \(F(A d P)\) acts in the upward direction. \(\therefore \quad F=A\left[-B\left(\dfrac{y A}{V}\right)\right]=-\dfrac{B A^{2}}{V} \cdot y\) \(\therefore \quad a=-\dfrac{B A^{2}}{m V} y\) comparing this with \(a=-\omega^{2} y\) \(\begin{aligned}\omega & =\sqrt{\dfrac{B A^{2}}{m V}} \\\Rightarrow \quad T & =\dfrac{2 \pi}{\omega}=2 \pi \sqrt{\dfrac{m V}{B A^{2}}}\end{aligned}\)
364325
A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency \(\omega\). The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time
1 For an amplitude of \(g^{2} / \omega^{2}\)
2 For an amplitude of \(g / \omega^{2}\)
3 At the mean position of the platform
4 At the highest position of the platform
Explanation:
The coin will leave contact when it is at the highest point and for that condition Maximum acceleration \(=\) Acceleration due to gravity \(\Rightarrow \omega^{2} A=g \quad \Rightarrow\) Amplitude \(A=\dfrac{g}{\omega^{2}}\).
PHXI14:OSCILLATIONS
364326
Consider a liquid which fills a uniform U-tube, as shown in fig up to a height \(h\). Find angular frequency of small oscillations of the liquid in the U-tube.
1 \(\sqrt{g / 2 h}\)
2 \(\sqrt{g / h}\)
3 \(\sqrt{2 g / h}\)
4 \(\sqrt{3 g / 2 h}\)
Explanation:
Suppose that the liquid is displaced slightly from equilibrium so that its level rises in one arm of the tube, while it is depressed in the second arm by the same amount, \(x\). If the density of the liquid is \(\rho\), then the total mechanical energy of the liquid column is The difference in the two levels is \(2 x\). So the column having length \(2 x\) produces restoring force on the entire liquid. \(\begin{aligned}& \rho A 2 x g=\rho A 2 h a \\& a=-\left(\dfrac{g}{h}\right) x \Rightarrow \omega=\sqrt{\dfrac{g}{h}}\end{aligned}\)
PHXI14:OSCILLATIONS
364327
A clock \(S\) is based on oscillation of a spring and a clock \(P\) is based on pendulum motion. Both clocks run at the same rate on the earth. On a planet having the same density as earth but twice the radius
1 \(S\) will run faster than \(P\)
2 \(P\) will run faster than \(S\)
3 both will run at the same rate as on the earth
4 both will run at the same rate which will be different from that on the earth
Explanation:
Acceleration due to gravity, \(g=\dfrac{G M}{R^{2}}=\dfrac{G \times \dfrac{4}{3} \pi R^{3} \rho}{R^{2}}=\dfrac{4}{3} \pi G \rho R\) \(\Rightarrow g \propto R\) For pendulum clock, \(g\) will increase on the planet so time period will decrease. But for spring clock, it will not change. Hence, \(P\) will run faster than \(S\)
PHXI14:OSCILLATIONS
364328
A particle under the action of a SHM has a period of \(3\,{\rm{sec}}\) and under the effect of another it has a period \(4\,{\rm{sec}}\). What will be its period under the combined action of both the SHM's in the same direction?
1 \(7\,\sec \)
2 \(5\,\sec \)
3 \(2.4\,\sec \)
4 \(0.4\,\sec \)
Explanation:
\(T=2 \pi \sqrt{\dfrac{m}{K}}\) or \(T \propto \dfrac{1}{\sqrt{K}}\) \(T\) has increased \(\dfrac{4}{3}\) times. \(K^{\prime}=\dfrac{9 K}{16}\) When both are combined \(\begin{aligned}K_{n e t} & =K+K_{1} \\& =\dfrac{25}{16} K\end{aligned}\) New time period will become \(\dfrac{4}{3}\) times of \(3\,{\rm{sec}}\) or \(2.4\,{\rm{sec}}\)
PHXI14:OSCILLATIONS
364329
An air chamber of volume \(V\), has a long neck of cross-sectional area A. A ball of mass \(m\) is fitted smoothly in the neck. The bulk modulus of air is \(B\). If the ball is pressed down slightly and released, the time period of its oscillation is
1 \(\pi \sqrt{\dfrac{2 m V}{B A^{2}}}\)
2 \(2 \pi \sqrt{\dfrac{m V}{2 B A^{2}}}\)
3 \(\dfrac{\pi}{2} \sqrt{\dfrac{m}{B A^{2}}}\)
4 \(2 \pi \sqrt{\dfrac{m V}{B A^{2}}}\)
Explanation:
Let the ball is pressed down by a small amount \(\mathrm{y}\), then the volume of air decreases by Ay. Then, excess pressure is \(d p=-B\left(\dfrac{d V}{V}\right)=-B\left(\dfrac{y A}{V}\right)\) A restoring force \(F(A d P)\) acts in the upward direction. \(\therefore \quad F=A\left[-B\left(\dfrac{y A}{V}\right)\right]=-\dfrac{B A^{2}}{V} \cdot y\) \(\therefore \quad a=-\dfrac{B A^{2}}{m V} y\) comparing this with \(a=-\omega^{2} y\) \(\begin{aligned}\omega & =\sqrt{\dfrac{B A^{2}}{m V}} \\\Rightarrow \quad T & =\dfrac{2 \pi}{\omega}=2 \pi \sqrt{\dfrac{m V}{B A^{2}}}\end{aligned}\)
364325
A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency \(\omega\). The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time
1 For an amplitude of \(g^{2} / \omega^{2}\)
2 For an amplitude of \(g / \omega^{2}\)
3 At the mean position of the platform
4 At the highest position of the platform
Explanation:
The coin will leave contact when it is at the highest point and for that condition Maximum acceleration \(=\) Acceleration due to gravity \(\Rightarrow \omega^{2} A=g \quad \Rightarrow\) Amplitude \(A=\dfrac{g}{\omega^{2}}\).
PHXI14:OSCILLATIONS
364326
Consider a liquid which fills a uniform U-tube, as shown in fig up to a height \(h\). Find angular frequency of small oscillations of the liquid in the U-tube.
1 \(\sqrt{g / 2 h}\)
2 \(\sqrt{g / h}\)
3 \(\sqrt{2 g / h}\)
4 \(\sqrt{3 g / 2 h}\)
Explanation:
Suppose that the liquid is displaced slightly from equilibrium so that its level rises in one arm of the tube, while it is depressed in the second arm by the same amount, \(x\). If the density of the liquid is \(\rho\), then the total mechanical energy of the liquid column is The difference in the two levels is \(2 x\). So the column having length \(2 x\) produces restoring force on the entire liquid. \(\begin{aligned}& \rho A 2 x g=\rho A 2 h a \\& a=-\left(\dfrac{g}{h}\right) x \Rightarrow \omega=\sqrt{\dfrac{g}{h}}\end{aligned}\)
PHXI14:OSCILLATIONS
364327
A clock \(S\) is based on oscillation of a spring and a clock \(P\) is based on pendulum motion. Both clocks run at the same rate on the earth. On a planet having the same density as earth but twice the radius
1 \(S\) will run faster than \(P\)
2 \(P\) will run faster than \(S\)
3 both will run at the same rate as on the earth
4 both will run at the same rate which will be different from that on the earth
Explanation:
Acceleration due to gravity, \(g=\dfrac{G M}{R^{2}}=\dfrac{G \times \dfrac{4}{3} \pi R^{3} \rho}{R^{2}}=\dfrac{4}{3} \pi G \rho R\) \(\Rightarrow g \propto R\) For pendulum clock, \(g\) will increase on the planet so time period will decrease. But for spring clock, it will not change. Hence, \(P\) will run faster than \(S\)
PHXI14:OSCILLATIONS
364328
A particle under the action of a SHM has a period of \(3\,{\rm{sec}}\) and under the effect of another it has a period \(4\,{\rm{sec}}\). What will be its period under the combined action of both the SHM's in the same direction?
1 \(7\,\sec \)
2 \(5\,\sec \)
3 \(2.4\,\sec \)
4 \(0.4\,\sec \)
Explanation:
\(T=2 \pi \sqrt{\dfrac{m}{K}}\) or \(T \propto \dfrac{1}{\sqrt{K}}\) \(T\) has increased \(\dfrac{4}{3}\) times. \(K^{\prime}=\dfrac{9 K}{16}\) When both are combined \(\begin{aligned}K_{n e t} & =K+K_{1} \\& =\dfrac{25}{16} K\end{aligned}\) New time period will become \(\dfrac{4}{3}\) times of \(3\,{\rm{sec}}\) or \(2.4\,{\rm{sec}}\)
PHXI14:OSCILLATIONS
364329
An air chamber of volume \(V\), has a long neck of cross-sectional area A. A ball of mass \(m\) is fitted smoothly in the neck. The bulk modulus of air is \(B\). If the ball is pressed down slightly and released, the time period of its oscillation is
1 \(\pi \sqrt{\dfrac{2 m V}{B A^{2}}}\)
2 \(2 \pi \sqrt{\dfrac{m V}{2 B A^{2}}}\)
3 \(\dfrac{\pi}{2} \sqrt{\dfrac{m}{B A^{2}}}\)
4 \(2 \pi \sqrt{\dfrac{m V}{B A^{2}}}\)
Explanation:
Let the ball is pressed down by a small amount \(\mathrm{y}\), then the volume of air decreases by Ay. Then, excess pressure is \(d p=-B\left(\dfrac{d V}{V}\right)=-B\left(\dfrac{y A}{V}\right)\) A restoring force \(F(A d P)\) acts in the upward direction. \(\therefore \quad F=A\left[-B\left(\dfrac{y A}{V}\right)\right]=-\dfrac{B A^{2}}{V} \cdot y\) \(\therefore \quad a=-\dfrac{B A^{2}}{m V} y\) comparing this with \(a=-\omega^{2} y\) \(\begin{aligned}\omega & =\sqrt{\dfrac{B A^{2}}{m V}} \\\Rightarrow \quad T & =\dfrac{2 \pi}{\omega}=2 \pi \sqrt{\dfrac{m V}{B A^{2}}}\end{aligned}\)
364325
A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency \(\omega\). The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time
1 For an amplitude of \(g^{2} / \omega^{2}\)
2 For an amplitude of \(g / \omega^{2}\)
3 At the mean position of the platform
4 At the highest position of the platform
Explanation:
The coin will leave contact when it is at the highest point and for that condition Maximum acceleration \(=\) Acceleration due to gravity \(\Rightarrow \omega^{2} A=g \quad \Rightarrow\) Amplitude \(A=\dfrac{g}{\omega^{2}}\).
PHXI14:OSCILLATIONS
364326
Consider a liquid which fills a uniform U-tube, as shown in fig up to a height \(h\). Find angular frequency of small oscillations of the liquid in the U-tube.
1 \(\sqrt{g / 2 h}\)
2 \(\sqrt{g / h}\)
3 \(\sqrt{2 g / h}\)
4 \(\sqrt{3 g / 2 h}\)
Explanation:
Suppose that the liquid is displaced slightly from equilibrium so that its level rises in one arm of the tube, while it is depressed in the second arm by the same amount, \(x\). If the density of the liquid is \(\rho\), then the total mechanical energy of the liquid column is The difference in the two levels is \(2 x\). So the column having length \(2 x\) produces restoring force on the entire liquid. \(\begin{aligned}& \rho A 2 x g=\rho A 2 h a \\& a=-\left(\dfrac{g}{h}\right) x \Rightarrow \omega=\sqrt{\dfrac{g}{h}}\end{aligned}\)
PHXI14:OSCILLATIONS
364327
A clock \(S\) is based on oscillation of a spring and a clock \(P\) is based on pendulum motion. Both clocks run at the same rate on the earth. On a planet having the same density as earth but twice the radius
1 \(S\) will run faster than \(P\)
2 \(P\) will run faster than \(S\)
3 both will run at the same rate as on the earth
4 both will run at the same rate which will be different from that on the earth
Explanation:
Acceleration due to gravity, \(g=\dfrac{G M}{R^{2}}=\dfrac{G \times \dfrac{4}{3} \pi R^{3} \rho}{R^{2}}=\dfrac{4}{3} \pi G \rho R\) \(\Rightarrow g \propto R\) For pendulum clock, \(g\) will increase on the planet so time period will decrease. But for spring clock, it will not change. Hence, \(P\) will run faster than \(S\)
PHXI14:OSCILLATIONS
364328
A particle under the action of a SHM has a period of \(3\,{\rm{sec}}\) and under the effect of another it has a period \(4\,{\rm{sec}}\). What will be its period under the combined action of both the SHM's in the same direction?
1 \(7\,\sec \)
2 \(5\,\sec \)
3 \(2.4\,\sec \)
4 \(0.4\,\sec \)
Explanation:
\(T=2 \pi \sqrt{\dfrac{m}{K}}\) or \(T \propto \dfrac{1}{\sqrt{K}}\) \(T\) has increased \(\dfrac{4}{3}\) times. \(K^{\prime}=\dfrac{9 K}{16}\) When both are combined \(\begin{aligned}K_{n e t} & =K+K_{1} \\& =\dfrac{25}{16} K\end{aligned}\) New time period will become \(\dfrac{4}{3}\) times of \(3\,{\rm{sec}}\) or \(2.4\,{\rm{sec}}\)
PHXI14:OSCILLATIONS
364329
An air chamber of volume \(V\), has a long neck of cross-sectional area A. A ball of mass \(m\) is fitted smoothly in the neck. The bulk modulus of air is \(B\). If the ball is pressed down slightly and released, the time period of its oscillation is
1 \(\pi \sqrt{\dfrac{2 m V}{B A^{2}}}\)
2 \(2 \pi \sqrt{\dfrac{m V}{2 B A^{2}}}\)
3 \(\dfrac{\pi}{2} \sqrt{\dfrac{m}{B A^{2}}}\)
4 \(2 \pi \sqrt{\dfrac{m V}{B A^{2}}}\)
Explanation:
Let the ball is pressed down by a small amount \(\mathrm{y}\), then the volume of air decreases by Ay. Then, excess pressure is \(d p=-B\left(\dfrac{d V}{V}\right)=-B\left(\dfrac{y A}{V}\right)\) A restoring force \(F(A d P)\) acts in the upward direction. \(\therefore \quad F=A\left[-B\left(\dfrac{y A}{V}\right)\right]=-\dfrac{B A^{2}}{V} \cdot y\) \(\therefore \quad a=-\dfrac{B A^{2}}{m V} y\) comparing this with \(a=-\omega^{2} y\) \(\begin{aligned}\omega & =\sqrt{\dfrac{B A^{2}}{m V}} \\\Rightarrow \quad T & =\dfrac{2 \pi}{\omega}=2 \pi \sqrt{\dfrac{m V}{B A^{2}}}\end{aligned}\)
364325
A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency \(\omega\). The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time
1 For an amplitude of \(g^{2} / \omega^{2}\)
2 For an amplitude of \(g / \omega^{2}\)
3 At the mean position of the platform
4 At the highest position of the platform
Explanation:
The coin will leave contact when it is at the highest point and for that condition Maximum acceleration \(=\) Acceleration due to gravity \(\Rightarrow \omega^{2} A=g \quad \Rightarrow\) Amplitude \(A=\dfrac{g}{\omega^{2}}\).
PHXI14:OSCILLATIONS
364326
Consider a liquid which fills a uniform U-tube, as shown in fig up to a height \(h\). Find angular frequency of small oscillations of the liquid in the U-tube.
1 \(\sqrt{g / 2 h}\)
2 \(\sqrt{g / h}\)
3 \(\sqrt{2 g / h}\)
4 \(\sqrt{3 g / 2 h}\)
Explanation:
Suppose that the liquid is displaced slightly from equilibrium so that its level rises in one arm of the tube, while it is depressed in the second arm by the same amount, \(x\). If the density of the liquid is \(\rho\), then the total mechanical energy of the liquid column is The difference in the two levels is \(2 x\). So the column having length \(2 x\) produces restoring force on the entire liquid. \(\begin{aligned}& \rho A 2 x g=\rho A 2 h a \\& a=-\left(\dfrac{g}{h}\right) x \Rightarrow \omega=\sqrt{\dfrac{g}{h}}\end{aligned}\)
PHXI14:OSCILLATIONS
364327
A clock \(S\) is based on oscillation of a spring and a clock \(P\) is based on pendulum motion. Both clocks run at the same rate on the earth. On a planet having the same density as earth but twice the radius
1 \(S\) will run faster than \(P\)
2 \(P\) will run faster than \(S\)
3 both will run at the same rate as on the earth
4 both will run at the same rate which will be different from that on the earth
Explanation:
Acceleration due to gravity, \(g=\dfrac{G M}{R^{2}}=\dfrac{G \times \dfrac{4}{3} \pi R^{3} \rho}{R^{2}}=\dfrac{4}{3} \pi G \rho R\) \(\Rightarrow g \propto R\) For pendulum clock, \(g\) will increase on the planet so time period will decrease. But for spring clock, it will not change. Hence, \(P\) will run faster than \(S\)
PHXI14:OSCILLATIONS
364328
A particle under the action of a SHM has a period of \(3\,{\rm{sec}}\) and under the effect of another it has a period \(4\,{\rm{sec}}\). What will be its period under the combined action of both the SHM's in the same direction?
1 \(7\,\sec \)
2 \(5\,\sec \)
3 \(2.4\,\sec \)
4 \(0.4\,\sec \)
Explanation:
\(T=2 \pi \sqrt{\dfrac{m}{K}}\) or \(T \propto \dfrac{1}{\sqrt{K}}\) \(T\) has increased \(\dfrac{4}{3}\) times. \(K^{\prime}=\dfrac{9 K}{16}\) When both are combined \(\begin{aligned}K_{n e t} & =K+K_{1} \\& =\dfrac{25}{16} K\end{aligned}\) New time period will become \(\dfrac{4}{3}\) times of \(3\,{\rm{sec}}\) or \(2.4\,{\rm{sec}}\)
PHXI14:OSCILLATIONS
364329
An air chamber of volume \(V\), has a long neck of cross-sectional area A. A ball of mass \(m\) is fitted smoothly in the neck. The bulk modulus of air is \(B\). If the ball is pressed down slightly and released, the time period of its oscillation is
1 \(\pi \sqrt{\dfrac{2 m V}{B A^{2}}}\)
2 \(2 \pi \sqrt{\dfrac{m V}{2 B A^{2}}}\)
3 \(\dfrac{\pi}{2} \sqrt{\dfrac{m}{B A^{2}}}\)
4 \(2 \pi \sqrt{\dfrac{m V}{B A^{2}}}\)
Explanation:
Let the ball is pressed down by a small amount \(\mathrm{y}\), then the volume of air decreases by Ay. Then, excess pressure is \(d p=-B\left(\dfrac{d V}{V}\right)=-B\left(\dfrac{y A}{V}\right)\) A restoring force \(F(A d P)\) acts in the upward direction. \(\therefore \quad F=A\left[-B\left(\dfrac{y A}{V}\right)\right]=-\dfrac{B A^{2}}{V} \cdot y\) \(\therefore \quad a=-\dfrac{B A^{2}}{m V} y\) comparing this with \(a=-\omega^{2} y\) \(\begin{aligned}\omega & =\sqrt{\dfrac{B A^{2}}{m V}} \\\Rightarrow \quad T & =\dfrac{2 \pi}{\omega}=2 \pi \sqrt{\dfrac{m V}{B A^{2}}}\end{aligned}\)
