Physics, asked by masoomaNaqvi, 3 months ago

A 0.12kg body undergoes SHM of amplitude 8.5cm and period 0.20s .
(a) what is the magnitude of the maximum force acting on it .
(b) if the oscillations are produced by a spring,what is the spring constant?​

Answers

Answered by sinhaprisha30
1

Explanation:

Answer

(a) The acceleration amplitude is related to the maximum force by Newton’s second law: F

max

=ma

m

. The acceleration amplitude is a

m

2

x

m

, where ω is the angular frequency (ω=2πf since there are 2π radians in one cycle).

The frequency is the reciprocal of the period: f=

T

1

=

0.20

1

=5.0Hz, so the angular frequency is ω=10π.

Therefore, F

max

=mω

2

x

m

=(0.12kg)(10πrad/s)

2

(0.085m) =10N.

(b) Using Eq. ω=

m

k

, we obtain

k=mω

2

=(0.12kg)(10πrad/s)

2

=1.21×10

2

N/m.

Answered by Anonymous
4

Answer:

(a) The acceleration amplitude is related to the maximum force by Newton’s second law:  Fmax​=mam​. The acceleration amplitude is  am​=ω2xm​, where  ω  is the angular frequency (ω=2πf  since there are  2π  radians in one cycle).

The frequency is the reciprocal of the period:  f=T1​ =0.201​=5.0Hz, so the angular frequency is  ω=10π.

Therefore,  Fmax​=mω2xm​ =(0.12kg)(10πrad/s)2(0.085m) =10N.

(b) Using Eq.  ω=mk​

​, we obtain  

k=mω2 =(0.12kg)(10πrad/s)2 =1.21×102N/m.

Explanation:

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