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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
10 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
10 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 (A2 - X2 )1/2 =W2X

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

PLZZZZZZZZZZ APPROVE IF SATISFIED

Kushagra Madhukar
askIITians Faculty 628 Points
3 years 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) = w√3
now acc at distance x is a= w2 x = w2
as |a| = |v|
w2 = w√3
w = √3
2π/T = √3
T = 2π/√3
 
Thanks and regards,
Kushagra

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