Advertisements
Advertisements
प्रश्न
A pendulum of 100 cm and another pendulum 4 cm long are oscillating at the same time. Calculate the ratio of their time periods.
उत्तर
l1 = 100 cm;
l2 = 4 cm
Let T1 and T2 be the time period of two pendulums.
`"T"_1/"T"_2=sqrt("l"_1/"l"_2)`
`"T"_1/"T"_2=sqrt(100/4)=sqrt(25/1)`
`"T"_1/"T"_2=5/1`
APPEARS IN
संबंधित प्रश्न
The time period of a pendulum clock is :
Compare the time periods of the two pendulums of lengths 1 m and 9 m.
Calculate the time period of a simple pendulum of length 1.44 m on the surface of the moon. The acceleration due to gravity on the surface of the moon is 1/6 the acceleration due to gravity on earth, [g = 9.8 ms−2]
Calculate the length of the second’s pendulum on the surface of the moon when acceleration due to gravity on the moon is 1.63 ms−2.
A second’s pendulum is taken on the surface of the moon where the acceleration due to gravity is l/6th of that of earth. Will the time period of pendulum remain the same or increase or decrease? Give a reason.
State the two ratios of a scale, which are not suitable for plotting points.
Moon has no atmosphere and acceleration due to gravity at its surface is one-sixth that at the earth's surface. When a pendulum is taken from earth to moon surface, its time period will:
What is a simple pendulum?
Name the factors on which the time period of a simple pendulum depends. Write the relation for the time period in terms of the above named factors.
