Question

Calcluate the time taken by the body performing s.h.m.to cover half amplitude starting from the extreme position with period second.

Answer 1

Answer:

If SHM starts from extreme , then it has initially phase of π/2.

Then its Equation will be :

x = A sin(ωt + π/2)

Calculation:

Now distance travelled is from A to A/2:

∴ A/2 = A sin(ωt + π/2)

=> ½ = cos(ωt)

=> π/3 = ωt

=> π/3 = {2π/T} × t

=> t = T/6

So final answer is T/6 , where T is the total time period is T

Answer 2

$\huge{\underline{\underline{\mathscr\green{Question\::}}}}$

Calcluate the time taken by the body performing s.h.m.to cover half amplitude starting from the extreme position with period second.

$\huge{\underline{\underline{\mathscr\pink{GIVEN\::}}}}$

•SHM IS STATING FROM THE EXTREME POSITION, THEN INITIAL PHASE OF π/2.

•X=A/2.

EQUATION IS,

X=Asin(wt+π/2)

$\huge{\underline{\underline{\mathscr\red{SOLUTION\::}}}}$

HERE,

$\implies$DISTANCE TRAVELLED FROM A TO A/2.

$\implies$.°.A/2=A SIN (wt +π/2)

$\implies$.°.½=COS (Wt)

$\implies$.°.π/3=WT

$\implies$.°.π/3=(2π/T) ×t. ( W VALUE 2π/T)

$\implies$.°.$\cancel{π}3$ =$\cancel{π}2/T$×t

$\implies$.°. 3×2/T=t

$\huge{\pink{\boxed{\boxed{\boxed{\purple{\underline{\underline{\green{\mathscr{.°. t=T/6}}}}}}}}}}$