A simple harmonic oscillator consists of a block of mass 2.2Kg attached to a spr

Question

A simple harmonic oscillator consists of a block of mass 2.2Kg attached to a spring of spring constant 370 N/m when t = 0.890 s, The position and velocity of the block are x= 0.112 m and v = 3.140 m/s (a) what is the amplitude of the oscillations? What were the (b) position and (c) velocity of the block at t = 0 s? A simple harmonic oscillator consists of a block of mass 2.2Kg attached to a spring of spring constant 370 N/m when t = 0.890 s, The position and velocity of the block are x= 0.112 m and v = 3.140 m/s (a) what is the amplitude of the oscillations? What were the (b) position and (c) velocity of the block at t = 0 s?

Explanation / Answer

Since,

k = m²

370 = (2.2)²

= 12.97 rad/s

Now, the equation for SHM is

y(t) = A cos (t + )

y'(t) = - A sin (t + )

y(0.89) = A cos (12.97(0.89) + ) = 0.112 => A cos(11.54+) = 0.112 .................(1)

y'(0.89) = - A(12.97) sin (12.97(0.89) + ) = 3.14 => - A sin(11.54+) = 0.242 ............(2)

(a) From the above two equations, A^2 = 0.112^2 + 0.242^2 = 0.071108

=> A = 0.267
Further,
cos(11.54+) = 0.112/0.267 = 0.419

= 53.69 degree = 0.94 rad.

so position and velocity at t = 0 are:

(b) y(t) = 0.267 cos (12.97t + 0.94)

y(0) = 0.267 cos (12.97(0) + 0.94)

y(0) = 0.157 m


(c) y'(t) = - 0.267(12.97) sin (12.97t + 0.94)

y'(0) = - 0.267(12.97) sin (12.97(0) + 0.94)

y'(0) = -2.80 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.