Advertisements
Advertisements
प्रश्न
A simple pendulum is constructed by hanging a heavy ball by a 5.0 m long string. It undergoes small oscillations. (a) How many oscillations does it make per second? (b) What will be the frequency if the system is taken on the moon where acceleration due to gravitation of the moon is 1.67 m/s2?
Advertisements
उत्तर
It is given that:
Length of the pendulum, l = 5 m
Acceleration due to gravity, g = 9.8 ms-2
Acceleration due to gravity at the moon, g' = 1.67 ms-2
(a) Time period \[\left( T \right)\] is given by,
\[T = 2\pi\sqrt{\frac{l}{g}}\]
\[= 2\pi\sqrt{\frac{5}{9 . 8}}\]
\[ = 2\pi\sqrt{0 . 510} = 2\pi \left( 0 . 71 \right) s\]
i.e. the body will take 2 \[\pi\](0.7) seconds to complete an oscillation.
Now, frequency \[\left( f \right)\]is given by,
\[f = \frac{1}{T}\]
\[\therefore f = \frac{1}{2\pi\left( 0 . 71 \right)} \]
\[ = \frac{0 . 70}{\pi} Hz\]
(b) Let
\[g'\] be the value of acceleration due to gravity at moon. Time period of simple pendulum at moon \[\left( T' \right)\],is given as:
\[T' = 2\pi\sqrt{\left( \frac{l}{g'} \right)}\]
On substituting the respective values in the above formula, we get:
\[ = \frac{1}{2\pi}\sqrt{\frac{1 . 67}{5}} = \frac{1}{2\pi}\left( 0 . 577 \right)\]
\[ = \frac{1}{2\pi\sqrt{3}} Hz\]
APPEARS IN
संबंधित प्रश्न
A particle executes S.H.M. with a period of 10 seconds. Find the time in which its potential energy will be half of its total energy.
A small creature moves with constant speed in a vertical circle on a bright day. Does its shadow formed by the sun on a horizontal plane move in a sample harmonic motion?
In measuring time period of a pendulum, it is advised to measure the time between consecutive passage through the mean position in the same direction. This is said to result in better accuracy than measuring time between consecutive passage through an extreme position. Explain.
A hollow sphere filled with water is used as the bob of a pendulum. Assume that the equation for simple pendulum is valid with the distance between the point of suspension and centre of mass of the bob acting as the effective length of the pendulum. If water slowly leaks out of the bob, how will the time period vary?
A block of known mass is suspended from a fixed support through a light spring. Can you find the time period of vertical oscillation only by measuring the extension of the spring when the block is in equilibrium?
The time period of a particle in simple harmonic motion is equal to the time between consecutive appearances of the particle at a particular point in its motion. This point is
The time period of a particle in simple harmonic motion is equal to the smallest time between the particle acquiring a particular velocity \[\vec{v}\] . The value of v is
The displacement of a particle in simple harmonic motion in one time period is
The displacement of a particle is given by \[\overrightarrow{r} = A\left( \overrightarrow{i} \cos\omega t + \overrightarrow{j} \sin\omega t \right) .\] The motion of the particle is
An object is released from rest. The time it takes to fall through a distance h and the speed of the object as it falls through this distance are measured with a pendulum clock. The entire apparatus is taken on the moon and the experiment is repeated
(a) the measured times are same
(b) the measured speeds are same
(c) the actual times in the fall are equal
(d) the actual speeds are equal
A pendulum clock giving correct time at a place where g = 9.800 m/s2 is taken to another place where it loses 24 seconds during 24 hours. Find the value of g at this new place.
A simple pendulum of length l is suspended through the ceiling of an elevator. Find the time period of small oscillations if the elevator (a) is going up with and acceleration a0(b) is going down with an acceleration a0 and (c) is moving with a uniform velocity.
A uniform rod of length l is suspended by an end and is made to undergo small oscillations. Find the length of the simple pendulum having the time period equal to that of the road.
A hollow sphere of radius 2 cm is attached to an 18 cm long thread to make a pendulum. Find the time period of oscillation of this pendulum. How does it differ from the time period calculated using the formula for a simple pendulum?
A particle is subjected to two simple harmonic motions given by x1 = 2.0 sin (100π t) and x2 = 2.0 sin (120 π t + π/3), where x is in centimeter and t in second. Find the displacement of the particle at (a) t = 0.0125, (b) t = 0.025.
A simple harmonic motion is given by, x = 2.4 sin ( 4πt). If distances are expressed in cm and time in seconds, the amplitude and frequency of S.H.M. are respectively,
Assume there are two identical simple pendulum clocks. Clock - 1 is placed on the earth and Clock - 2 is placed on a space station located at a height h above the earth's surface. Clock - 1 and Clock - 2 operate at time periods 4 s and 6 s respectively. Then the value of h is ______.
(consider the radius of earth RE = 6400 km and g on earth 10 m/s2)
If x = `5 sin (pi t + pi/3) m` represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are ______.
