Question

A 170 g block attached to a spring with spring constant 2.1 N/m oscillates horizontally on a frictionless table. Its velocity is 19 cm/s when x0 = -4.0 cm .

A) What is the amplitude of oscillation?

B) What is the block's maximum acceleration?

C) What is the block's position when the acceleration is maximum?

D) What is the speed of the block when x1 = 3.3 cm ?

Answer 1

Given,

mass of the block, m = 170 g = 0.170 kg

Spring constant, k = 2.1 N/m

A)

Equation of SHM is given by

x = A sint

Now,

=>  

                = = 3.515

Now,

v =

=> 0.19 = 3.515

=> = 0.19 / 3.515 = 0.054

=> A² - 0.0016 = (0.054)² = 0.002916

=> A² = 0.002916 + 0.0016 = 0.004516

=> A = = 0.0672 m

or Amplitude is 6.72 cm

b)

Maximum accelearion, a_max = ²A

                                               = 12.353 * 0.0672 = 0.83 m/s²

                                               = 83 cm/s²

c)

Position of the block is at +6.72 cm or -6.72 cm

d)

v =

=

=

= = 3.515 * 0.0585

= 0.2056 m/s = 20.56 cm/s

or velocity is 20.56 cm/s