1 Time period of a simple pendulum depends on amplitude.
2 Time shown by a spring watch varies with acceleration due to gravity $\mathrm{g}$.
3 In a simple pendulum time period varies linearly with the length of the pendulum.
4 The graph between length of the pendulum and time period is a parabola.
Explanation:
D Time period of a SHM is given by $\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$ It does not depends upon the amplitude. Time period shown by spring watch is given by $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$ It does not depends upon the acceleration due to gravity From equation (i) $\mathrm{T}=\sqrt{l}$ Squares on both side in (i) we get - $\mathrm{T}^{2}=4 \pi^{2} \frac{l}{\mathrm{~g}}$ $l=4 \pi^{2} \mathrm{gT}^{2}$ It is an equation of an parabola. Hence, the graph between length of the pendulum and time period is a parabola.
J and K CET- 2006
Oscillations
140569
Two pendulums begin to swing simultaneously. If the ratio of the frequency of oscillations of the two is $7: 8$, then the ratio of lengths of the two pendulums will be
1 $7: 8$
2 $8: 7$
3 $49: 64$
4 $64: 49$
Explanation:
D Given as, Ratio of the frequency of oscillations $\left(\mathrm{f}_{1}: \mathrm{f}_{2}\right)=7: 8$ We know that, Frequency $(\mathrm{f})=\frac{1}{\text { Time }-\operatorname{Period}(\mathrm{T})} \Rightarrow \frac{1}{2 \pi \sqrt{\frac{l}{\mathrm{~g}}}}$ $\mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{g}}{l}}$ $\mathrm{f} \propto \frac{1}{\sqrt{l}}$ $\therefore \frac{\mathrm{f}_{1}}{\mathrm{f}_{2}}=\sqrt{\frac{l_{2}}{l_{1}}}$ So, $\quad \sqrt{\frac{l_{2}}{l_{1}}}=\frac{7}{8}$ $\frac{l_{2}}{l_{1}}=\frac{49}{64}$ $\frac{l_{1}}{l_{2}}=\frac{64}{49}$ $l_{1}: l_{2} \Rightarrow 64: 49$
J and K CET- 2005
Oscillations
140570
Two simple pendulums of lengths $1.44 \mathrm{~m}$ and 1 m start swinging together. After how many vibrations will they again start swinging together?
1 5 oscillations of smaller pendulum
2 6 oscillations of smaller pendulum
3 4 oscillations of bigger pendulum
4 6 oscillations of bigger pendulum
Explanation:
B Given data, length, $l_{1}=1.44 \mathrm{~m}$ and $l_{2}=1 \mathrm{~m}$ We know that, Time-period $(\mathrm{T})=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$ $\mathrm{T}_{1}=2 \pi \sqrt{\frac{l_{1}}{\mathrm{~g}}} \Rightarrow 2 \pi \sqrt{\frac{1.44}{\mathrm{~g}}}$ $\mathrm{T}_{2}=2 \pi \sqrt{\frac{l_{2}}{\mathrm{~g}}} \Rightarrow 2 \pi \sqrt{\frac{1}{\mathrm{~g}}}$ From equation (i) and equation (ii) $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{2 \pi \sqrt{\frac{1.44}{\mathrm{~g}}}}{2 \pi \sqrt{\frac{1}{\mathrm{~g}}}}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\sqrt{\frac{1.44}{\mathrm{~g}} \times \frac{\mathrm{g}}{1}}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\sqrt{\frac{144}{100}}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{12}{10}$ $5 \mathrm{~T}_{1}=6 \mathrm{~T}_{2}$ $6 \text { Oscillation of smaller pendulum }=5 \text { oscillation of }$ $\text { bigger pendulum }$
J and K CET- 2005
Oscillations
140571
The time period of a simple pendulum when it is made to oscillate on the surface of moon
1 increases
2 decreases
3 remains unchanged
4 becomes infinite
Explanation:
A We know that, for simple pendulum at earth $(\mathrm{T})=2 \pi \sqrt{\frac{l}{g}}$ On moon, $g$ is much smaller compared to $g$ on earth $\because \mathrm{g}=\mathrm{g}$ at earth $\mathrm{g}_{\text {moon }}=\frac{\mathrm{g}_{\text {earth }}}{6}$ So, at moon $\left(\mathrm{T}^{\prime}\right)=2 \pi \sqrt{\frac{l}{g / 6}} \Rightarrow 2 \pi \sqrt{\frac{6 l}{g}}$ $\mathrm{T}^{\prime}=\sqrt{6} \times \mathrm{T}$ At the surface of moon, $g$ decrease hence time period increases.
J and K CET- 2004
Oscillations
140573
The time period of a simple pendulum of length $9.8 \mathrm{~m}$ is
1 $0.159 \mathrm{sec}$
2 $3.14 \mathrm{sec}$
3 $5.6 \mathrm{sec}$
4 $6.28 \mathrm{sec}$
Explanation:
D Given that, length $(l)=9.8 \mathrm{~m}$ $\mathrm{~g}=9.8 \mathrm{~m} / \mathrm{s}^{2}$ we know that, $\mathrm{T}=2 \pi \sqrt{\frac{l}{g}}$ $\mathrm{~T}=2 \times 3.14 \sqrt{\frac{9.8}{9.8}}$ $\mathrm{~T}=6.28 \mathrm{sec}$
1 Time period of a simple pendulum depends on amplitude.
2 Time shown by a spring watch varies with acceleration due to gravity $\mathrm{g}$.
3 In a simple pendulum time period varies linearly with the length of the pendulum.
4 The graph between length of the pendulum and time period is a parabola.
Explanation:
D Time period of a SHM is given by $\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$ It does not depends upon the amplitude. Time period shown by spring watch is given by $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$ It does not depends upon the acceleration due to gravity From equation (i) $\mathrm{T}=\sqrt{l}$ Squares on both side in (i) we get - $\mathrm{T}^{2}=4 \pi^{2} \frac{l}{\mathrm{~g}}$ $l=4 \pi^{2} \mathrm{gT}^{2}$ It is an equation of an parabola. Hence, the graph between length of the pendulum and time period is a parabola.
J and K CET- 2006
Oscillations
140569
Two pendulums begin to swing simultaneously. If the ratio of the frequency of oscillations of the two is $7: 8$, then the ratio of lengths of the two pendulums will be
1 $7: 8$
2 $8: 7$
3 $49: 64$
4 $64: 49$
Explanation:
D Given as, Ratio of the frequency of oscillations $\left(\mathrm{f}_{1}: \mathrm{f}_{2}\right)=7: 8$ We know that, Frequency $(\mathrm{f})=\frac{1}{\text { Time }-\operatorname{Period}(\mathrm{T})} \Rightarrow \frac{1}{2 \pi \sqrt{\frac{l}{\mathrm{~g}}}}$ $\mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{g}}{l}}$ $\mathrm{f} \propto \frac{1}{\sqrt{l}}$ $\therefore \frac{\mathrm{f}_{1}}{\mathrm{f}_{2}}=\sqrt{\frac{l_{2}}{l_{1}}}$ So, $\quad \sqrt{\frac{l_{2}}{l_{1}}}=\frac{7}{8}$ $\frac{l_{2}}{l_{1}}=\frac{49}{64}$ $\frac{l_{1}}{l_{2}}=\frac{64}{49}$ $l_{1}: l_{2} \Rightarrow 64: 49$
J and K CET- 2005
Oscillations
140570
Two simple pendulums of lengths $1.44 \mathrm{~m}$ and 1 m start swinging together. After how many vibrations will they again start swinging together?
1 5 oscillations of smaller pendulum
2 6 oscillations of smaller pendulum
3 4 oscillations of bigger pendulum
4 6 oscillations of bigger pendulum
Explanation:
B Given data, length, $l_{1}=1.44 \mathrm{~m}$ and $l_{2}=1 \mathrm{~m}$ We know that, Time-period $(\mathrm{T})=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$ $\mathrm{T}_{1}=2 \pi \sqrt{\frac{l_{1}}{\mathrm{~g}}} \Rightarrow 2 \pi \sqrt{\frac{1.44}{\mathrm{~g}}}$ $\mathrm{T}_{2}=2 \pi \sqrt{\frac{l_{2}}{\mathrm{~g}}} \Rightarrow 2 \pi \sqrt{\frac{1}{\mathrm{~g}}}$ From equation (i) and equation (ii) $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{2 \pi \sqrt{\frac{1.44}{\mathrm{~g}}}}{2 \pi \sqrt{\frac{1}{\mathrm{~g}}}}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\sqrt{\frac{1.44}{\mathrm{~g}} \times \frac{\mathrm{g}}{1}}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\sqrt{\frac{144}{100}}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{12}{10}$ $5 \mathrm{~T}_{1}=6 \mathrm{~T}_{2}$ $6 \text { Oscillation of smaller pendulum }=5 \text { oscillation of }$ $\text { bigger pendulum }$
J and K CET- 2005
Oscillations
140571
The time period of a simple pendulum when it is made to oscillate on the surface of moon
1 increases
2 decreases
3 remains unchanged
4 becomes infinite
Explanation:
A We know that, for simple pendulum at earth $(\mathrm{T})=2 \pi \sqrt{\frac{l}{g}}$ On moon, $g$ is much smaller compared to $g$ on earth $\because \mathrm{g}=\mathrm{g}$ at earth $\mathrm{g}_{\text {moon }}=\frac{\mathrm{g}_{\text {earth }}}{6}$ So, at moon $\left(\mathrm{T}^{\prime}\right)=2 \pi \sqrt{\frac{l}{g / 6}} \Rightarrow 2 \pi \sqrt{\frac{6 l}{g}}$ $\mathrm{T}^{\prime}=\sqrt{6} \times \mathrm{T}$ At the surface of moon, $g$ decrease hence time period increases.
J and K CET- 2004
Oscillations
140573
The time period of a simple pendulum of length $9.8 \mathrm{~m}$ is
1 $0.159 \mathrm{sec}$
2 $3.14 \mathrm{sec}$
3 $5.6 \mathrm{sec}$
4 $6.28 \mathrm{sec}$
Explanation:
D Given that, length $(l)=9.8 \mathrm{~m}$ $\mathrm{~g}=9.8 \mathrm{~m} / \mathrm{s}^{2}$ we know that, $\mathrm{T}=2 \pi \sqrt{\frac{l}{g}}$ $\mathrm{~T}=2 \times 3.14 \sqrt{\frac{9.8}{9.8}}$ $\mathrm{~T}=6.28 \mathrm{sec}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Oscillations
140568
Choose the correct statement.
1 Time period of a simple pendulum depends on amplitude.
2 Time shown by a spring watch varies with acceleration due to gravity $\mathrm{g}$.
3 In a simple pendulum time period varies linearly with the length of the pendulum.
4 The graph between length of the pendulum and time period is a parabola.
Explanation:
D Time period of a SHM is given by $\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$ It does not depends upon the amplitude. Time period shown by spring watch is given by $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$ It does not depends upon the acceleration due to gravity From equation (i) $\mathrm{T}=\sqrt{l}$ Squares on both side in (i) we get - $\mathrm{T}^{2}=4 \pi^{2} \frac{l}{\mathrm{~g}}$ $l=4 \pi^{2} \mathrm{gT}^{2}$ It is an equation of an parabola. Hence, the graph between length of the pendulum and time period is a parabola.
J and K CET- 2006
Oscillations
140569
Two pendulums begin to swing simultaneously. If the ratio of the frequency of oscillations of the two is $7: 8$, then the ratio of lengths of the two pendulums will be
1 $7: 8$
2 $8: 7$
3 $49: 64$
4 $64: 49$
Explanation:
D Given as, Ratio of the frequency of oscillations $\left(\mathrm{f}_{1}: \mathrm{f}_{2}\right)=7: 8$ We know that, Frequency $(\mathrm{f})=\frac{1}{\text { Time }-\operatorname{Period}(\mathrm{T})} \Rightarrow \frac{1}{2 \pi \sqrt{\frac{l}{\mathrm{~g}}}}$ $\mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{g}}{l}}$ $\mathrm{f} \propto \frac{1}{\sqrt{l}}$ $\therefore \frac{\mathrm{f}_{1}}{\mathrm{f}_{2}}=\sqrt{\frac{l_{2}}{l_{1}}}$ So, $\quad \sqrt{\frac{l_{2}}{l_{1}}}=\frac{7}{8}$ $\frac{l_{2}}{l_{1}}=\frac{49}{64}$ $\frac{l_{1}}{l_{2}}=\frac{64}{49}$ $l_{1}: l_{2} \Rightarrow 64: 49$
J and K CET- 2005
Oscillations
140570
Two simple pendulums of lengths $1.44 \mathrm{~m}$ and 1 m start swinging together. After how many vibrations will they again start swinging together?
1 5 oscillations of smaller pendulum
2 6 oscillations of smaller pendulum
3 4 oscillations of bigger pendulum
4 6 oscillations of bigger pendulum
Explanation:
B Given data, length, $l_{1}=1.44 \mathrm{~m}$ and $l_{2}=1 \mathrm{~m}$ We know that, Time-period $(\mathrm{T})=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$ $\mathrm{T}_{1}=2 \pi \sqrt{\frac{l_{1}}{\mathrm{~g}}} \Rightarrow 2 \pi \sqrt{\frac{1.44}{\mathrm{~g}}}$ $\mathrm{T}_{2}=2 \pi \sqrt{\frac{l_{2}}{\mathrm{~g}}} \Rightarrow 2 \pi \sqrt{\frac{1}{\mathrm{~g}}}$ From equation (i) and equation (ii) $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{2 \pi \sqrt{\frac{1.44}{\mathrm{~g}}}}{2 \pi \sqrt{\frac{1}{\mathrm{~g}}}}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\sqrt{\frac{1.44}{\mathrm{~g}} \times \frac{\mathrm{g}}{1}}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\sqrt{\frac{144}{100}}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{12}{10}$ $5 \mathrm{~T}_{1}=6 \mathrm{~T}_{2}$ $6 \text { Oscillation of smaller pendulum }=5 \text { oscillation of }$ $\text { bigger pendulum }$
J and K CET- 2005
Oscillations
140571
The time period of a simple pendulum when it is made to oscillate on the surface of moon
1 increases
2 decreases
3 remains unchanged
4 becomes infinite
Explanation:
A We know that, for simple pendulum at earth $(\mathrm{T})=2 \pi \sqrt{\frac{l}{g}}$ On moon, $g$ is much smaller compared to $g$ on earth $\because \mathrm{g}=\mathrm{g}$ at earth $\mathrm{g}_{\text {moon }}=\frac{\mathrm{g}_{\text {earth }}}{6}$ So, at moon $\left(\mathrm{T}^{\prime}\right)=2 \pi \sqrt{\frac{l}{g / 6}} \Rightarrow 2 \pi \sqrt{\frac{6 l}{g}}$ $\mathrm{T}^{\prime}=\sqrt{6} \times \mathrm{T}$ At the surface of moon, $g$ decrease hence time period increases.
J and K CET- 2004
Oscillations
140573
The time period of a simple pendulum of length $9.8 \mathrm{~m}$ is
1 $0.159 \mathrm{sec}$
2 $3.14 \mathrm{sec}$
3 $5.6 \mathrm{sec}$
4 $6.28 \mathrm{sec}$
Explanation:
D Given that, length $(l)=9.8 \mathrm{~m}$ $\mathrm{~g}=9.8 \mathrm{~m} / \mathrm{s}^{2}$ we know that, $\mathrm{T}=2 \pi \sqrt{\frac{l}{g}}$ $\mathrm{~T}=2 \times 3.14 \sqrt{\frac{9.8}{9.8}}$ $\mathrm{~T}=6.28 \mathrm{sec}$
1 Time period of a simple pendulum depends on amplitude.
2 Time shown by a spring watch varies with acceleration due to gravity $\mathrm{g}$.
3 In a simple pendulum time period varies linearly with the length of the pendulum.
4 The graph between length of the pendulum and time period is a parabola.
Explanation:
D Time period of a SHM is given by $\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$ It does not depends upon the amplitude. Time period shown by spring watch is given by $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$ It does not depends upon the acceleration due to gravity From equation (i) $\mathrm{T}=\sqrt{l}$ Squares on both side in (i) we get - $\mathrm{T}^{2}=4 \pi^{2} \frac{l}{\mathrm{~g}}$ $l=4 \pi^{2} \mathrm{gT}^{2}$ It is an equation of an parabola. Hence, the graph between length of the pendulum and time period is a parabola.
J and K CET- 2006
Oscillations
140569
Two pendulums begin to swing simultaneously. If the ratio of the frequency of oscillations of the two is $7: 8$, then the ratio of lengths of the two pendulums will be
1 $7: 8$
2 $8: 7$
3 $49: 64$
4 $64: 49$
Explanation:
D Given as, Ratio of the frequency of oscillations $\left(\mathrm{f}_{1}: \mathrm{f}_{2}\right)=7: 8$ We know that, Frequency $(\mathrm{f})=\frac{1}{\text { Time }-\operatorname{Period}(\mathrm{T})} \Rightarrow \frac{1}{2 \pi \sqrt{\frac{l}{\mathrm{~g}}}}$ $\mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{g}}{l}}$ $\mathrm{f} \propto \frac{1}{\sqrt{l}}$ $\therefore \frac{\mathrm{f}_{1}}{\mathrm{f}_{2}}=\sqrt{\frac{l_{2}}{l_{1}}}$ So, $\quad \sqrt{\frac{l_{2}}{l_{1}}}=\frac{7}{8}$ $\frac{l_{2}}{l_{1}}=\frac{49}{64}$ $\frac{l_{1}}{l_{2}}=\frac{64}{49}$ $l_{1}: l_{2} \Rightarrow 64: 49$
J and K CET- 2005
Oscillations
140570
Two simple pendulums of lengths $1.44 \mathrm{~m}$ and 1 m start swinging together. After how many vibrations will they again start swinging together?
1 5 oscillations of smaller pendulum
2 6 oscillations of smaller pendulum
3 4 oscillations of bigger pendulum
4 6 oscillations of bigger pendulum
Explanation:
B Given data, length, $l_{1}=1.44 \mathrm{~m}$ and $l_{2}=1 \mathrm{~m}$ We know that, Time-period $(\mathrm{T})=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$ $\mathrm{T}_{1}=2 \pi \sqrt{\frac{l_{1}}{\mathrm{~g}}} \Rightarrow 2 \pi \sqrt{\frac{1.44}{\mathrm{~g}}}$ $\mathrm{T}_{2}=2 \pi \sqrt{\frac{l_{2}}{\mathrm{~g}}} \Rightarrow 2 \pi \sqrt{\frac{1}{\mathrm{~g}}}$ From equation (i) and equation (ii) $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{2 \pi \sqrt{\frac{1.44}{\mathrm{~g}}}}{2 \pi \sqrt{\frac{1}{\mathrm{~g}}}}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\sqrt{\frac{1.44}{\mathrm{~g}} \times \frac{\mathrm{g}}{1}}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\sqrt{\frac{144}{100}}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{12}{10}$ $5 \mathrm{~T}_{1}=6 \mathrm{~T}_{2}$ $6 \text { Oscillation of smaller pendulum }=5 \text { oscillation of }$ $\text { bigger pendulum }$
J and K CET- 2005
Oscillations
140571
The time period of a simple pendulum when it is made to oscillate on the surface of moon
1 increases
2 decreases
3 remains unchanged
4 becomes infinite
Explanation:
A We know that, for simple pendulum at earth $(\mathrm{T})=2 \pi \sqrt{\frac{l}{g}}$ On moon, $g$ is much smaller compared to $g$ on earth $\because \mathrm{g}=\mathrm{g}$ at earth $\mathrm{g}_{\text {moon }}=\frac{\mathrm{g}_{\text {earth }}}{6}$ So, at moon $\left(\mathrm{T}^{\prime}\right)=2 \pi \sqrt{\frac{l}{g / 6}} \Rightarrow 2 \pi \sqrt{\frac{6 l}{g}}$ $\mathrm{T}^{\prime}=\sqrt{6} \times \mathrm{T}$ At the surface of moon, $g$ decrease hence time period increases.
J and K CET- 2004
Oscillations
140573
The time period of a simple pendulum of length $9.8 \mathrm{~m}$ is
1 $0.159 \mathrm{sec}$
2 $3.14 \mathrm{sec}$
3 $5.6 \mathrm{sec}$
4 $6.28 \mathrm{sec}$
Explanation:
D Given that, length $(l)=9.8 \mathrm{~m}$ $\mathrm{~g}=9.8 \mathrm{~m} / \mathrm{s}^{2}$ we know that, $\mathrm{T}=2 \pi \sqrt{\frac{l}{g}}$ $\mathrm{~T}=2 \times 3.14 \sqrt{\frac{9.8}{9.8}}$ $\mathrm{~T}=6.28 \mathrm{sec}$
1 Time period of a simple pendulum depends on amplitude.
2 Time shown by a spring watch varies with acceleration due to gravity $\mathrm{g}$.
3 In a simple pendulum time period varies linearly with the length of the pendulum.
4 The graph between length of the pendulum and time period is a parabola.
Explanation:
D Time period of a SHM is given by $\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$ It does not depends upon the amplitude. Time period shown by spring watch is given by $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$ It does not depends upon the acceleration due to gravity From equation (i) $\mathrm{T}=\sqrt{l}$ Squares on both side in (i) we get - $\mathrm{T}^{2}=4 \pi^{2} \frac{l}{\mathrm{~g}}$ $l=4 \pi^{2} \mathrm{gT}^{2}$ It is an equation of an parabola. Hence, the graph between length of the pendulum and time period is a parabola.
J and K CET- 2006
Oscillations
140569
Two pendulums begin to swing simultaneously. If the ratio of the frequency of oscillations of the two is $7: 8$, then the ratio of lengths of the two pendulums will be
1 $7: 8$
2 $8: 7$
3 $49: 64$
4 $64: 49$
Explanation:
D Given as, Ratio of the frequency of oscillations $\left(\mathrm{f}_{1}: \mathrm{f}_{2}\right)=7: 8$ We know that, Frequency $(\mathrm{f})=\frac{1}{\text { Time }-\operatorname{Period}(\mathrm{T})} \Rightarrow \frac{1}{2 \pi \sqrt{\frac{l}{\mathrm{~g}}}}$ $\mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{g}}{l}}$ $\mathrm{f} \propto \frac{1}{\sqrt{l}}$ $\therefore \frac{\mathrm{f}_{1}}{\mathrm{f}_{2}}=\sqrt{\frac{l_{2}}{l_{1}}}$ So, $\quad \sqrt{\frac{l_{2}}{l_{1}}}=\frac{7}{8}$ $\frac{l_{2}}{l_{1}}=\frac{49}{64}$ $\frac{l_{1}}{l_{2}}=\frac{64}{49}$ $l_{1}: l_{2} \Rightarrow 64: 49$
J and K CET- 2005
Oscillations
140570
Two simple pendulums of lengths $1.44 \mathrm{~m}$ and 1 m start swinging together. After how many vibrations will they again start swinging together?
1 5 oscillations of smaller pendulum
2 6 oscillations of smaller pendulum
3 4 oscillations of bigger pendulum
4 6 oscillations of bigger pendulum
Explanation:
B Given data, length, $l_{1}=1.44 \mathrm{~m}$ and $l_{2}=1 \mathrm{~m}$ We know that, Time-period $(\mathrm{T})=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$ $\mathrm{T}_{1}=2 \pi \sqrt{\frac{l_{1}}{\mathrm{~g}}} \Rightarrow 2 \pi \sqrt{\frac{1.44}{\mathrm{~g}}}$ $\mathrm{T}_{2}=2 \pi \sqrt{\frac{l_{2}}{\mathrm{~g}}} \Rightarrow 2 \pi \sqrt{\frac{1}{\mathrm{~g}}}$ From equation (i) and equation (ii) $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{2 \pi \sqrt{\frac{1.44}{\mathrm{~g}}}}{2 \pi \sqrt{\frac{1}{\mathrm{~g}}}}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\sqrt{\frac{1.44}{\mathrm{~g}} \times \frac{\mathrm{g}}{1}}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\sqrt{\frac{144}{100}}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{12}{10}$ $5 \mathrm{~T}_{1}=6 \mathrm{~T}_{2}$ $6 \text { Oscillation of smaller pendulum }=5 \text { oscillation of }$ $\text { bigger pendulum }$
J and K CET- 2005
Oscillations
140571
The time period of a simple pendulum when it is made to oscillate on the surface of moon
1 increases
2 decreases
3 remains unchanged
4 becomes infinite
Explanation:
A We know that, for simple pendulum at earth $(\mathrm{T})=2 \pi \sqrt{\frac{l}{g}}$ On moon, $g$ is much smaller compared to $g$ on earth $\because \mathrm{g}=\mathrm{g}$ at earth $\mathrm{g}_{\text {moon }}=\frac{\mathrm{g}_{\text {earth }}}{6}$ So, at moon $\left(\mathrm{T}^{\prime}\right)=2 \pi \sqrt{\frac{l}{g / 6}} \Rightarrow 2 \pi \sqrt{\frac{6 l}{g}}$ $\mathrm{T}^{\prime}=\sqrt{6} \times \mathrm{T}$ At the surface of moon, $g$ decrease hence time period increases.
J and K CET- 2004
Oscillations
140573
The time period of a simple pendulum of length $9.8 \mathrm{~m}$ is
1 $0.159 \mathrm{sec}$
2 $3.14 \mathrm{sec}$
3 $5.6 \mathrm{sec}$
4 $6.28 \mathrm{sec}$
Explanation:
D Given that, length $(l)=9.8 \mathrm{~m}$ $\mathrm{~g}=9.8 \mathrm{~m} / \mathrm{s}^{2}$ we know that, $\mathrm{T}=2 \pi \sqrt{\frac{l}{g}}$ $\mathrm{~T}=2 \times 3.14 \sqrt{\frac{9.8}{9.8}}$ $\mathrm{~T}=6.28 \mathrm{sec}$