#### Thank you for registering.

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

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0
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

# a particle executes linear shmbwith an amplitude if 2 cm.when the particle is at 1cm from the mean position the magnitude of its velocity is equal to that of its acceleration.Then its time period is...??

8 years ago

velocity at distance x in shm is given as

v= w sqrt(A2-x2)

so A = 2 x =1

v = w sqrt(4-1) = wsqrt3

now acc at distance x is a= w2 x = w2

as mag a = mag v

w2 = wsqrt3

w= sqrt3

2pi/T= sqrt3

T = 2pi/sqrt3

8 years ago

V = W (A2 - X2 )1/2

A=W2X

WHERE ''W'' IS ANGULAR FREQUENCY LET ''T'' BE TIME PERIOD T=2PIE/W

X=0.01m AND A=0.02m

ACCORDING TO THE GIVEN CONDITION

V=A

THERE FORE

W (A- X)1/2 =W2X

PLZZZZZZZZZZ APPROVE IF SATISFIED Kushagra Madhukar
9 months ago
Dear student,

Velocity at distance x in shm is given as
v = w(A2 – x2)
Here,
A = 2
x =1
Hence, v = w (4-1) = w3
now acc at distance x is a= w2 x = w2
as |a| = |v|
w2 = w3
w = 3
2π/T = 3
T = 2π/3

Thanks and regards,
Kushagra