Question

Q1) A 1.2 kg block sliding on a horizontal frictionless surface is attached to a horizontal spring with k =480 N/m. Let x be the displacement of the block from the position at which the spring is upstretched. At t = 0 the block passes through x = 0 with a speed of 5.2 m/s in the positive x direction. What are the (a) frequency and (b) amplitude of the block’s motion? (c) Write an expression for x as a function of time

Answer 1

Step-by-step explanation:

given,

mass of block = 1.2 kg

spring constant (k) = 480 N/m

a) Frequency

`f = (1)/(2\pi )\sqrt{(k)/(m)}`

`f = (1)/(2\pi )\sqrt{(480)/(1.2)}`

f = 3.183 Hz

b) The position function

x = A cos (ωt + ∅)

at t = 0 , x = 0

0 = A cos ∅

cos ∅ = 0

∅ = π/2

velocity at t= 0 s

`v = (dx)/(dt)`

v = Aωsin (ωt + ∅)

`v = A\omega sin ( + (\pi )/(2))`

`5.2 = A\sqrt{(480)/(1.2)}`

A = 0.26 m

c)
`x = 0.26 cos (\sqrt{(480)/(1.2)}t + (\pi )/(2))`

`x = 0.26 cos( 20 t +(\pi )/(2))`