मराठी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान इयत्ता ११

A Simple Pendulum of Length 40 Cm is Taken Inside a Deep Mine. Assume for the Time Being that the Mine is 1600 Km Deep.

Advertisements
Advertisements

प्रश्न

A simple pendulum of length 40 cm is taken inside a deep mine. Assume for the time being that the mine is 1600 km deep. Calculate the time period of the pendulum there. Radius of the earth = 6400 km.

बेरीज
Advertisements

उत्तर

It is given that:
Length of the pendulum, l = 40 cm = 0.4 m
Radius of the earth, R = 6400 km
Acceleration due to gravity on the earth's surface, `g = 9.8 "ms"^(- 2)`

Let

\[g'\] be the acceleration due to gravity at a depth of 1600 km from the surface of the earth. 
Its value is given by,

\[g' = g\left( 1 - \frac{d}{R} \right)\] \[\text{where  d  is  the  depth  from  the  earth  surfce, }\] \[\text { R  is  the  radius  of  earth,   and }\] 

\[\text {g  is  acceleration  due  to  gravity .}\]

\[\text{ On  substituting  the  respective  values,   we  get: }\] \[  g' = 9 . 8\left( 1 - \frac{1600}{6400} \right)\] 

\[= 9 . 8\left( 1 - \frac{1}{4} \right)\] 

\[= 9 . 8 \times \left( \frac{3}{4} \right) = 7 . 35   {\text{ms}}^{- 2}\]

Time period is given as,

\[T = 2\pi\sqrt{\left( \frac{l}{g'} \right)}\]

\[\Rightarrow T = 2\pi\sqrt{\left( \frac{0 . 4}{7 . 35} \right)}\] 

\[ \Rightarrow T = 2 \times 3 . 14 \times 0 . 23\] 

\[             = 1 . 465 \approx 1 . 47  s\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 12: Simple Harmonics Motion - Exercise [पृष्ठ २५५]

APPEARS IN

एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
पाठ 12 Simple Harmonics Motion
Exercise | Q 40 | पृष्ठ २५५

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

The average displacement over a period of S.H.M. is ______.

(A = amplitude of S.H.M.)


A body of mass 1 kg is made to oscillate on a spring of force constant 16 N/m. Calculate:

a) Angular frequency

b) frequency of vibration.


Can simple harmonic motion take place in a non-inertial frame? If yes, should the ratio of the force applied with the displacement be constant?


The energy of system in simple harmonic motion is given by \[E = \frac{1}{2}m \omega^2 A^2 .\] Which of the following two statements is more appropriate?
(A) The energy is increased because the amplitude is increased.
(B) The amplitude is increased because the energy is increased.


Can a pendulum clock be used in an earth-satellite?


A student says that he had applied a force \[F = - k\sqrt{x}\] on a particle and the particle moved in simple harmonic motion. He refuses to tell whether k is a constant or not. Assume that he was worked only with positive x and no other force acted on the particle.


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

 

A particle moves on the X-axis according to the equation x = A + B sin ωt. The motion is simple harmonic with amplitude


The average energy in one time period in simple harmonic motion is


The motion of a torsional pendulum is
(a) periodic
(b) oscillatory
(c) simple harmonic
(d) angular simple harmonic


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 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 particle is subjected to two simple harmonic motions of same time period in the same direction. The amplitude of the first motion is 3.0 cm and that of the second is 4.0 cm. Find the resultant amplitude if the phase difference between the motions is (a) 0°, (b) 60°, (c) 90°.


Three simple harmonic motions of equal amplitude A and equal time periods in the same direction combine. The phase of the second motion is 60° ahead of the first and the phase of the third motion is 60° ahead of the second. Find the amplitude of the resultant motion.


Define the frequency of simple harmonic motion.


Write short notes on two springs connected in series.


Consider a simple pendulum of length l = 0.9 m which is properly placed on a trolley rolling down on a inclined plane which is at θ = 45° with the horizontal. Assuming that the inclined plane is frictionless, calculate the time period of oscillation of the simple pendulum.


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, 


The displacement of a particle varies with time according to the relation y = a sin ωt + b cos ωt.


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)


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×