Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

Solution of a particle is performing simple harmonic motion along x-axis with amplitude 4cm and time period 1.2 sec . The minimum time taken by the particle to move from x=+2 cm to x=+4 cm and back again is given by?

Solution of a particle is performing simple harmonic motion along x-axis with amplitude 4cm and time period 1.2 sec . The minimum time taken by the particle to move from x=+2 cm to x=+4 cm and back again is given by?

Grade:12th Pass

2 Answers

Narasimha rao
16 Points
9 years ago

given

T=1.2 sec.

the min time to move from x=2 to x=4 is asked i.e.,

the time taken by the particle to move form half of its amplitude to its max amplitude

let the eqn be x=Asinwt

the time taken to go from its mean position to extreme position is T/4

time taken to cover half of its amplitude be T1

at t=0,x=0,v=Aw

at x=A/2,the time T1 is

A/2=AsinwT1,wT1=pi/6,T1=pi/6w=T/12

so time taken to move from A/2 to A is

t=T/4-T/12=T/6=1.2/6=0.2 seconds

at extreme position the velocity is zero,acceleration negative

at mid position x=A/2,

we can work it out as above and get t=T/12,t=1.2/12=.1

total time=.2+.1=.3 seconds

 

APURV GOEL
39 Points
9 years ago

basically the question is time taken for going from A/2 to A and come back where A is amplitude, which comes out to be T/3 where T is time period so ans is 1.2/3 = 0.4sec

PLEASE APPROVE.

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free