Simple Harmonic Motion Typical Problem

[thunderhadron]

Hi friends the problem is -

https://fbcdn-sphotos-d-a.akamaihd.net/hphotos-ak-prn1/s480x480/60061_2656517749482_1458399262_n.jpg

Attempt -

As per the problem states,

The net force on the particle will be ,

https://fbcdn-sphotos-a-a.akamaihd.net/hphotos-ak-ash4/s480x480/222254_2656522429599_1932558900_n.jpg

which is giving no way to get the answer as per the question.

The correct answer of this problem is option (B), (∏/3) as per the question.

Please friends help me in solving this Problem.

Thank you all in advance.

 

[TSny]

I think you need to interpret the notation carefully. A and B are points along a line that are considered fixed. A and B do not have numerical values, they are just labels for the two points. P is also a label for some arbitrary point on the line. The notation $\overline{PA}$ denotes a vector pointing in the direction from point P to point A with magnitude equal to the distance between P and A. Similarly for $\overline{PB}$.

A force of $2\overline{PA}$ denotes a force pointing in the direction from P to A and magnitude 2 times the distance between P and A. The "2" should be thought of has having units so the force comes out in units of N.

 

[thunderhadron]

I think you need to interpret the notation carefully. A and B are points along a line that are considered fixed. A and B do not have numerical values, they are just labels for the two points. P is also a label for some arbitrary point on the line. The notation $\overline{PA}$ denotes a vector pointing in the direction from point P to point A with magnitude equal to the distance between P and A. Similarly for $\overline{PB}$.

A force of $2\overline{PA}$ denotes a force pointing in the direction from P to A and magnitude 2 times the distance between P and A. The "2" should be thought of has having units so the force comes out in units of N.

Thanks a lot TSny I got its answer too.