Question

If theta denotes the anular displacement and alpha the angular amplitude of a simple pendulum oscillating in a vertical plane and m the mass of the bob,the tension in the string is more than mgcos theta
(A)never possible
(b)always possible
(c)only when theta =alpha
(d)only when theta 14 years ago

Answer 1

What is the tension in the string when the angular displacement is theta?

Answer 2

mgcos theta i suppose

Answer 3

Yes, hence the answer is (a) never.

Answer 4

actually you mean that the the tension adjust itself with varying theta .so at any part of its motion the tension is always balanced by mg cos theta .is it?

Answer 5

There's only one force acting on the bob and hence only one force acting on the string. The force the bob exerts on the string is always mg cos(theta). Hence the statement
"the tension in the string is more than mgcos theta"
is always false.

Answer 6

but why is alpa and theta described in the qn .can u pls distinguish them

Answer 7

Alpha is the maximum angular displacement.

If the question had asked about the statement
"the tension in the string is more than mg.cos(alpha)"
the answer would have been: yes, this is true, except when the bob is at the end of its swing, either side, and
\[ |\theta| < \alpha

Answer 8

\[ |\theta| < \alpha \]

Answer 9

because for a maximum angular displacement \( \alpha \) where
\[ 0 < \alpha < \pi/2 \]
we have that for any \( \theta \in (-\alpha, \alpha) \),
\[ \cos \theta < \cos \alpha \]
because cos is a even function which has a local maximum at 0.

Answer 10

angular displacement theta can be any value depending on the position of bob.but angular ampliude is always constant is it?

Answer 11

Yes, it is the maximum angle of displacement the pendulum achieves at either end of its swing: it oscillates between \( \alpha \) and \( -\alpha \).

Answer 12

but if theta less than alpha the cos alpha ,then cos theta will be greater than cos alpha isnt't because cos is a decreasing fn

Answer 13

i think u made a typo there

Answer 14

Yes, sorry I meant to say
\[ \cos \theta > \cos \alpha \]

Answer 15

:P

Answer 16

thanks

Answer 17

sure thing

Answer 18

i have an interesting qn on gravitation can u try out?

Answer 19

post it as a new question

Answer 20

What is the supposed answer and what is the explanation.

Answer 21

(C)

Answer 22

my teacher told we need to consider mv^2/r .but i dont know why

Answer 23

srry (D)

Answer 24

Curious. I think then the argument is this:
when the mass is below the maximum angular displacement it has kinetic energy equal to
\[ KE = \Delta PE = PE(\alpha) - PE(\theta) = mgr(\cos \theta - \cos \alpha) \]
Now as
\[ KE = \frac{1}{2}mv^2 \] and the the centripetal force is
\[ F_c = \frac{mv^2}{r} = \frac{2KE}{r} \] this means that
\[ F_c = 2mg(\cos \theta - \cos \alpha) \]

Answer 25

why we consider centripeta force?

Answer 26

Now this is the tension in the rope and this quantity is greater than more than
mg.cos(theta) if
\[ 2(\cos \theta - \cos \alpha) > \cos \theta \]
i.e.,
\[ \cos \theta > 2 \cos \alpha \]
That doesn't quite get us there either.

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You should press him for a worked solution. I don't see how D is the answer.

Answer 27

look this