IIT JEE , AIEEE Forum - Physics , Mechanics

Koel

A simple pendulum is oscillating without damping.When the displacement of the bob is less than maximum, its acc. vector a is correctly shown in :(plz explain ur ans.)

pink_ele

first option

becoz motion is circular so force is towards center

rini

the correct ans should be B.

because when bob is at this position, it has a component of centripital acceleration and one of gravitation.. hense resultanant should be in this direction..

a5hw1n_5

the pendulum executes simple harmonic motion, hence its accleration is always towards the mean position.

therefore the answer should be B

elessar_iitkgp

The correct answer is the left hand side bottom one.
At the position of the bob shown, it has two acceleration components:
The tangential acceleration gsin

along the tangent and oriented toward left
And the centripetal accln. v²/L directed inwards.
Their vector sum will be something as shown in the left hand side bottom one.

umang

The answer is c , i.e. , the bottom left . The acc. has two components - the first is the centripetal one , directed inwards , as shown in first option , and the second is tangantial component , as shown in second option . Hence , the net accelaration will be vector sum of these two components , and it could be something like that shown in option 3 .

govindsuku

the ans is b ......coz when the bob is at the extreme its tangential acc ( opp to disp - gsinX) is max..........and as it is reaching the extreme its velocity is almost 0......so centripetal acc ( mv^2/r) = 0.......therefore.....only tangential acc is present and tht is represented by option b

priyesh

the answer is c) the bottom left one because the acceleration will be resultant of mean position vector i.e gsin(theta) & centripetal one i.e T - mgcos(theta) = mv^2/L

rini

but the tangential acc is also a component of gravitational acc and hence only g and centripital acc should be taken into account..
the ans should be B.

rakesh_89

ans b

bcoz 2 forces r acting on particle one is mg other T . T is always canceled by mgcos

hence net force, accn is gsin

as shown in fig.

rakesh_89

rectification

ans c

the a force should be acting toward the centre c.p.f. hence resultant accn is

(gcos

^2+(v^2/r)^2) in dir shown in fig.

catch_arnnie

there are two accelerations on the bob ::

1) tangential ---> due to g

2) radial ----> due to centripetal force

the resultant of both the acelerations is correctly shown in the bottom-left figure.

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