Recent content by vishwesh

Thanks for the reply. I think I got it. The max output should be 1 when ##\omega \rightarrow \infty##?

Thanks for the reply. But, I didn't get you. I am familiar with that formula. I understand that we use that formula to substitute ##\cfrac{|V_2|}{|V_1|} = \cfrac{1}{\sqrt{2}}## but how does it explain the ##\cfrac{1}{2\sqrt{2}}## part ? Thanks again.

Homework Statement *(Problems attached as images)* For first problem: ##\cfrac{|V_2|}{|V_1|} = \cfrac{1}{\sqrt{4+(\omega RC)^2}}## For second problem: ##\cfrac{|V_2|}{|V_1|} = \cfrac{\sqrt{1+(\omega RC)^2}}{\sqrt{4 + (\omega RC)^2}}## I have to calculate half power frequency for both...

Thanks for the reply. I am also stuck with what the question expects me to do. As for the force in x-direction, can you please tell me which other force will act ?

Thanks for the reply. I will wait for some other responses.

Homework Statement A mass m hangs on a spring of constant k. In the position of static equilibrium the length of the spring is l. If the mass is drawn sideways and then released,the ensuing motion will be a combination of (a) pendulum swings and (b) extension and compression of the spring...

Superb! Thanks a lot for your help.

Hi! Whenever you face problems with probability questions of this sort, try to build a venn diagram (like the one I have attached). Use the venn diagram to calculate n(A∩B) and then Pr(A∩B). Then, you will have to use your third formula because here a condition is given ---> he/she has a job...

Thanks for the reply. Net moment of inertia, ##I_{net} = I_{1} + I_{2} = \cfrac{1}{2} MR^2 + (\cfrac{1}{2}MR^2 + Ml^2)## ##\implies I_{net} = MR^2 + Ml^2## Plugging in the values in the formula: ##\omega = \sqrt{\cfrac{mgl}{I}} = \sqrt{\cfrac{(2M)g(\cfrac{l}{2})}{MR^2 + Ml^2}}## ## \implies...

Homework Statement Problem statement - [/B] Klepner and Kolenkow 6.15 : A pendulum is made of two disks each of mass M and radius R separated by a massless rod. One of the disks is pivoted through its center by a small pin. The disks hang in the same plane and their centres are a distance l...