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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...??

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...?? 

Grade:12

3 Answers

Har Simrat Singh
42 Points
9 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

ankitesh gupta
63 Points
9 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

FIND THE VALUE OF ''W'' AND GET YOUR ANSWER 

PLZZZZZZZZZZ APPROVE IF SATISFIED

Kushagra Madhukar
askIITians Faculty 629 Points
one year ago
Dear student,
Please find the solution to your problem below.
 
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

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