Question

Few questions I have doubts on:

Answer 1

1. Two simple pendulums of length 1 m and 4m are both given small displacement in same direction at the same instant. They will be again in phase after the short pendulum has completed number of oscillations=.....

Answer 2

\[2T_2=T_1\]
\[2 \pi f_1=2(2 \pi f_2)\]

Answer 3

Now what to do?

Answer 4

2 periods for the smaller one = 1 period for the longer one

so the answer is 2.

Answer 5

Oh is that all? I was confused.

Answer 6

Unless what they call "oscillation" is not a full period but half a period.

In that case, the answer would be 4.

Answer 7

A bullet of mass 10 g and speed 500 m/s is fired into a door and gets embedded at exactly center of the door. The door is 1 m wide and weighs 12kg. It is hinged at one end and rotates in a vertical axis. The angular speed of door just after bullet embeds it will be?

Answer 8

Use conservation of angular momentum.
0.010kg x 500 x 0.5m = I x omega

with I = m L²/3 = 12 x 1² /3 = 4 kg.m²

omega = 0.6 rad/s

Answer 9

So for the bullet door system angular momentum is conserved always?

Answer 10

Yes because there is no net torque acting and the collision time is infinitesimal.

Answer 11

Orbits of a particle moving in a circle are such that perimeter of orbit=nlambda
where lambda is de broglie wavelength. For a charged particle moving in a plane perpendicular to mag field radius of nth orbit will be proportional to?

Answer 12

Ok I know how to do this qn. my doubt is why the general eqn \[r \alpha n^2 \]
Does not hold here.

Answer 13

I do not understand. Is it alpha between r and n² ?

Answer 14

I mean r proportional to n^2 for an electron moving around nucleus.

Answer 15

It must be different here, as the particle is not orbiting an atom with an 1/r² force but is moving in a magnetic field.
I'm not sure. I'll think about it. See you.

Answer 16

ok bye. Thanks again!