मराठी
महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

Find the change in length of a second’s pendulum, if the acceleration due to gravity at the place changes from 9.75 m/s2 to 9.8 m/s2.

Advertisements
Advertisements

प्रश्न

Find the change in length of a second’s pendulum, if the acceleration due to gravity at the place changes from 9.75 m/s2 to 9.8 m/s2.

बेरीज
Advertisements

उत्तर

Data: gf =9.75 m/s2 , g2 = 9.8 m/s2

Length of a seconds pendulum, L = `"g"/π^2`

∴ L1 = `"g"_1/"π"^2=9.75/9.872` = 0.9876 m

and L2 = `"g"_2/π^2=9.8/9.872` = 0.9927 m

Find the change in length

ΔL = L2 − L1

= 0.9927 − 0.9876

= 0.0051 m = 5.1 mm

∴ The length of the second's pendulum must be increased from 0.9876 m to 0.9927 m, i.e., by 0.0051 m.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 5: Oscillations - Exercises [पृष्ठ १३०]

APPEARS IN

बालभारती Physics [English] Standard 12 Maharashtra State Board
पाठ 5 Oscillations
Exercises | Q 15 | पृष्ठ १३०

संबंधित प्रश्‍न

Acceleration of a particle executing S.H.M. at its mean position.


Two identical wires of substances 'P' and 'Q ' are subjected to equal stretching force along the length. If the elongation of 'Q' is more than that of 'P', then ______.


The displacement of a particle from its mean position (in metre) is given by, y = 0.2 sin(10 πt + 1.5π) cos(10 πt + 1.5π).

The motion of particle is ____________.


A body performing a simple harmonic motion has potential energy 'P1' at displacement 'x1' Its potential energy is 'P2' at displacement 'x2'. The potential energy 'P' at displacement (x1 + x2) is ________.


The phase difference between the instantaneous velocity and acceleration of a particle executing S.H.M is ____________.


If 'α' and 'β' are the maximum velocity and maximum acceleration respectively, of a particle performing linear simple harmonic motion, then the path length of the particle is _______.


Which of the following represents the acceleration versus displacement graph of SHM?


The displacement of a particle is 'y' = 2 sin `[(pit)/2 + phi]`, where 'y' is cm and 't' in second. What is the maximum acceleration of the particle executing simple harmonic motion? 

(Φ = phase difference)


The length of the second's pendulum is decreased by 0.3 cm when it is shifted from place A to place B. If the acceleration due to gravity at place A is 981 cm/s2, the acceleration due to gravity at place B is ______ (Take π2 = 10)


A simple pendulum of length 'L' is suspended from a roof of a trolley. A trolley moves in horizontal direction with an acceleration 'a'. What would be the period of oscillation of a simple pendulum?

(g is acceleration due to gravity)


The bob of a simple pendulum is released at time t = 0 from a position of small angular displacement. Its linear displacement is ______.

(l = length of simple pendulum and g = acceleration due to gravity, A = amplitude of S.H.M.)


The displacement of a particle in S.H.M. is x = A cos `(omegat+pi/6).` Its speed will be maximum at time ______.


The displacement of a particle is represented by the equation `y = 3 cos (pi/4 - 2ωt)`. The motion of the particle is ______.


In figure, a particle is placed at the highest point A of a smooth sphere of radius r. It is given slight push and it leaves the sphere at B, at a depth h vertically below A, such that h is equal to ______.


A particle of mass 5 kg moves in a circle of radius 20 cm. Its linear speed at a time t is given by v = 4t, t is in the second and v is in ms-1. Find the net force acting on the particle at t = 0.5 s.


A body of mass 0.5 kg travels in a straight line with velocity v = ax3/2 where a = 5 m–1/2s–1. The change in kinetic energy during its displacement from x = 0 to x = 2 m is ______.


In the given figure, a = 15 m/s2 represents the total acceleration of a particle moving in the clockwise direction on a circle of radius R = 2.5 m at a given instant of time. The speed of the particle is ______.


Calculate the velocity of a particle performing S.H.M. after 1 second, if its displacement is given by x = `5sin((pit)/3)`m.


For a particle performing circular motion, when is its angular acceleration directed opposite to its angular velocity?


State the expression for the total energy of SHM in terms of acceleration.


Which one of the following is not a characteristics of SHM?


A particle executing SHM has velocities v1 and v2 when it is at distance x1 and x2 from the centre of the path. Show that the time period is given by `T=2pisqrt((x_2^2-x_1^2)/(v_1^2-v_2^2))`


A particle performs simple harmonic motion with amplitude A. Its speed is tripled at the instant that it is at a distance `(2 A)/3` from equilibrium position. The new amplitude of the motion is ______.


A pendulum is performing simple harmonic motion. The acceleration of the bob is 20 cm s−2 at a distance of 5 cm from mean position. The time period of oscillation is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×