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dedylja [7]
3 years ago
9

A block is attached to a spring, with spring constant k, which is attached to a wall. it is initially moved to the left a distan

ce d (at point a and then released from rest, where the block undergoes harmonic motion. the floor is frictionless.
Physics
1 answer:
Ghella [55]3 years ago
3 0

Answer:

x(t) = d*cos ( wt )

w = √(k/m)

Explanation:

Given:-

- The mass of block = m

- The spring constant = k

- The initial displacement = xi = d

Find:-

- The expression for displacement (x) as function of time (t).

Solution:-

- Consider the block as system which is initially displaced with amount (x = d) to left and then released from rest over a frictionless surface and undergoes SHM. There is only one force acting on the block i.e restoring force of the spring F = -kx in opposite direction to the motion.

- We apply the Newton's equation of motion in horizontal direction.

                             F = ma

                             -kx = ma

                             -kx = mx''

                              mx'' + kx = 0

- Solve the Auxiliary equation for the ODE above:

                              ms^2 + k = 0

                              s^2 + (k/m) = 0

                              s = +/- √(k/m) i = +/- w i

- The complementary solution for complex roots is:

                              x(t) = [ A*cos ( wt ) + B*sin ( wt ) ]

- The given initial conditions are:

                              x(0) = d

                              d = [ A*cos ( 0 ) + B*sin ( 0 ) ]

                              d = A

                              x'(0) = 0

                              x'(t) = -Aw*sin (wt) + Bw*cos(wt)

                              0 = -Aw*sin (0) + Bw*cos(0)

                              B = 0

- The required displacement-time relationship for SHM:

                               x(t) = d*cos ( wt )

                               w = √(k/m)

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An insulating cup contains 200 grams of water at 25 ∘C. Some ice cubes at 0 ∘C is placed in the water. The system comes to equil
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Answer:

The amount of ice added in gram is 32.77g

Explanation:

This problem bothers on the heat capacity of materials

Given data

Mass of water Mw= 200g

Temperature of water θw= 25°c

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Equilibrium Temperature θe= 12°c

Mass of ice Mi=???

The specific heat of ice Ci= 2090 J/(kg ∘C)

specific heat of water Cw = 4186 J/(kg ∘C)

latent heat of the ice to water transition Li= 3.33 x10^5 J/kg

heat heat loss by water = heat gained by ice

N/B let us understand something, heat gained by ice is in two phases

Heat require to melt ice at 0°C to water at 0°C

And the heat required to take water from 0°C to equilibrium temperature

Hence

MwCwΔθ=MiLi +MiCiΔθ

Substituting our data we have

200*4186*(25-12)=Mi*3.3x10^5+

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