Question

A graph of the variation of kinetic energy of an object of mass 5kg with position x from equilibrium position l , -x=0,4 to x =0,4 and wx= 4. Find period of oscillation
Calculate velocity of the object at a displacement of 0,25 m from the equilibrium position and maximum acceleration of the object.

Answer 1

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Answer 2

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Answer 3

To find the period of oscillation, we can use the formula:

T = 2π/ω

where T is the period and ω is the angular frequency.
Given that wx = 4, we can find ω using the formula:

ω = sqrt(k/m)

where k is the spring constant and m is the mass of the object.
Since the graph shows the variation of kinetic energy, we can use the equation:

KE = (1/2)mv^2

where KE is the kinetic energy and v is the velocity of the object.

To calculate the velocity of the object at a displacement of 0.25 m from the equilibrium position, we can use the formula:

PE = (1/2)kx^2

Since the given graph is of kinetic energy, we can use the equation:

KE = PE

Thus, we have:

(1/2)mv^2 = (1/2)kx^2

Since m = 5 kg, x = 0.25 m, and we know that wx = 4, we can rearrange the equation to solve for v:

v^2 = (kx^2)/m

v^2 = (wx^2)/m

v^2 = (4 * (0.25)^2) / 5

v^2 = 0.04

v = sqrt(0.04)

v = 0.2 m/s

To calculate the maximum acceleration, we can use the equation:

a = -ω^2x

Since wx = 4 and x = 0.25, we have:

a = -4^2 * 0.25

a = -4 m/s^2

Therefore, the velocity of the object at a displacement of 0.25 m is 0.2 m/s and the maximum acceleration of the object is -4 m/s^2.

Answer 4

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Answer 5

Alright, if you have any questions or need assistance with anything, feel free to ask!