An ideal spring is lying horizontally on a frictionless surface. One end of the spring is attached to a wall. The other end is attached to a moveable block that has a mass of 5 kg. The block is pulled so that the spring stretches from its equilibrium position by 0.65 m. Then the block is released (from rest), and as a result the system oscillates with a frequency of 0.40 Hz (that's 0.40 rev/sec) Find: a) the acceleratiuon of the block when the spring is stretched by 0.28 m.
b) the maximum force magnitude exerted by the spring on the block.
c) the oscillation frequency of a 2.5 kg blcok under the same circumstances (i. e. with the same spring and initial displacement).

a) a = - 1.76 m / s²,  b)   F = 20.5 N, c)  w = 3.55 rad / s

Explanation:

a) a simple harmonic motion is described by the expression

          x = A cos (wt + Ф)

in this case they give us the frequency

         w = 2π f

         w = 2π 0.40

         w = 2.51 rad / s

as the maximum elongation is 0.65 m this corresponds to the amplitude of the movement

         A = 0.65 m

to find the phase constant (Ф) we use the initial condition that for t = 0 v = 0 and x = A, we substitute

          A = A cos (0+ Ф)

         cos Ф = 1

         Ф = 0

the resulting equation is

         x = 0.65 cos (2.51 t)

Let's find the time it takes to get to x = 0.28 m

         0.28 = 0.65 cos 2.51 t

          2.51 t = cos-1 (0.28 / 0.65)

remember angles are in radians

          t = 1.1254 / 2.51

          t = 0.448 s

the acceleration is

         a = -A w² cos wt

we subtitle

         a = - 0.65 2.51² cos (2.51  0.448)

         a = - 1.76 m / s²

b) the maximum acceleration occurs when the cosine is ±1

        a = A w²

        a = 0.65  2.51²

        a = 4.10 m / s²

Let's use Newton's second law

        F = m a

        F = 5 4.1

        F = 20.5 N

c) The angular velocity is given by

          w² = k / m

let's find the spring constant

          k = m w²

          k = 5 2.51²

          k = 31.5 N / m

therefore if the block is exchanged for another with mass m'= 2.5 kg

         w = √(31.5 / 2.5)

         w = 3.55 rad / s

emily ruth's annual premium is $1058.50

step-by-step explanation:

emily ruth is a female of 27 years age purchasing 20 year endowment insurance with face value $25000.

from the given table,

for females, the age of a male 3 years younger should be used.

so, we have to check for 24 year male for 20 year endowment.

it is 42.34 for every $1000 face value.

for a face value of $25000,

it is 25 * 42.34 = $1058.50

∴ emily ruth's annual premium is $1058.50