Review Conceptual Example 8 before starting this problem. A block is attached to a horizontal spring and oscillates back and forth on a frictionless horizontal surface at a frequency of 3.96 Hz. The amplitude of the motion is 5.95 x 10-2 m. At the point where the block has its maximum speed, it suddenly splits into two identical parts, only one part remaining attached to the spring. (a) What is the amplitude and (b) the frequency of the simple harmonic motion that exists after the block splits

a)  A' =  0.345  m,  b)  f = 2,800 Hz

Explanation:

b) The angular velocity of a simple harmonic motion is

        w =

angular velocity and frequency are related

        w = 2π f

we substitute

        f = 1 /2π   √k/m

indicates that the initial frequency value f = 3.96 Hz

in this case the mass is reduced by half

       m ’= m / 2

we substitute

       f = 2π

       f = √1/2    (2π √k/m)

       f = 1 /√2  3.96

       f = 2,800 Hz

a) The amplitude of the movement is defined by the value of the initial depalzamienot before an external force that initiates the movement.

When the block is divided into two parts of equal masses as if it were exploding, for which we can use the conservation of moment

initial instant. Right before the division

        p₀ = (m₁ + m₁) v

final instant. Right after the split

        p_f = m₁ v '

        p₀ = p_f

        (2 m₁) v = m₁ v ’

        v ’= 2v

At this point we can use conservation of energy for the system with only half the block.

Starting point. Where the block divides

         Em₀o = K = ½ m v'²

Final point. Point of maximum elongation

          Em_f = Ke = ½ k A²

how energy is conserved

         Em₀ = Em_f

         ½ m’ v’² = ½ k A’²

we substitute the previous expressions

         ½ m/2 (2v)² = ½ k A’²

         A’² = 2  m v² / k                       (1)

Let's use the conservation of energy with the initial conditions, before dividing the block

          ½ m v2 = ½ k A2

          A² = mv² / k = 5.95 10⁻² m²

we substitute in 1

         A'² = 2 A²

          A ’²=  2 5.95 10⁻²

          A ’²= 11.9 10⁻² m

          A' =  0.345  m

statement #3 on andre"s list is false.

force is based upon mass and acceleration.the statement is not always true.

it can be explained from newton's second law of motion.

as per newton's second law of motion the rate of change of momentum is directly proportional to the applied force and takes place along the direction of force.

mathematically it can be written as -

                                              ⇒      

here   k is the proportionality constant whose value is 1.

                                                            [a is the acceleration=dv/dt]

hence we see that force not only depends on mass and acceleration but also on velocity and rate of change of mass. force f will depend on mass and acceleration only when mass is constant.

hence the statement that force is based upon mass and acceleration is false.