Mechanics> linear-shm...

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...??

Paramjeet singh , 12 Years ago

Grade 12

3 Answers

Har Simrat Singh

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

Last Activity: 12 Years ago

ankitesh gupta

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

Last Activity: 12 Years ago

Kushagra Madhukar

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

Last Activity: 5 Years ago

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Mechanics> linear-shm...

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...??

Paramjeet singh , 12 Years ago

Har Simrat Singh

ankitesh gupta

Kushagra Madhukar

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