If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released, it will oscillate. The object is displaced a distance 0.120 m from its equilibrium position and released with zero initial speed. Then after a time 0.800 s, its displacement is found to be a distance 0.120 m on the opposite side, and it has passed the equilibrium position once during this interval. a) Find the amplitude. b) Find the period. c) Find the frequency."

a)0.120m

b)1.60s

c)0.625Hz

Explanation:

a)In this problem, initial displacement A=0.120m was displaced and released with zero initial speed.

The amplitude is the maximum displacement from equilibrium and it was maximum at the moment t=0.(∴The object goes from x=+A to x=-A and back to complete one cycle)

So the amplitude will be equal to the initial displacement A=0.120m

Therefore, the amplitude is 0.120m

b) movement from +A( max positive displacement) to -A (max negative displacement) took place during half the period of SHM

therefore,

0.800 s= T/2   ---> by further solving for T

T= 0.800 x 2

T=1.6 s

The period is 1.6 s

c) As we know that frequency is inversely proportional to the time period.

To find frequency:

f=1/T           (∴T=1.6)

f=1/1.6

f=0.625Hz

therefore, the frequency is 0.625 Hz

a) 0.120m, b) 1.6s, c) 0.625Hz

Explanation:

a)

From the question, we can see that the maximum displacement from the mean position is 0.120m, this value is the amplitude of the motion.

Hence A = 0.120m

b)

From the point of release to the opposite end, the object has completed half an oscillation ( assuming it went back to it point of release, then we have a complete oscillation), hence the period of the motion is half which is also equals to the time taken to move from the point of release to the opposite end (0.800s)

0.800 = T/2

Where T is the period of motion.

By cross multiplying, we have that

T = 0.800×2

T = 1.6s

c)

There is an inverse relationship between frequency and period.

Where f = 1/T

f = 1/0.16 = 0.625 Hz

-4 on

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j