A simple harmonic oscillator consists of a block of mass 1.00 kg attached to a s

Question

A simple harmonic oscillator consists of a block of mass 1.00 kg attached to a spring of spring constant 220 N/m. When t = 2.50 s, the position and velocity of the block are x = 0.146 m and v = 3.360 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 1.00 kg attached to a spring of spring constant 220 N/m. When t = 2.50 s, the position and velocity of the block are x = 0.146 m and v = 3.360 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 1.00 kg attached to a spring of spring constant 220 N/m. When t = 2.50 s, the position and velocity of the block are x = 0.146 m and v = 3.360 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

here,

mass of block , m = 1 kg

spring constant , k = 220 N/m

at t = 2.5 s

velocity , v = 3.36 m/s

position , x = 0.146 m

(a)

let the amplitude of oscillation be A

using conservation of energy

0.5 * kx^2 + 0.5 * mv^2 = 0.5 * k * A^2

0.5 * 220 * 0.146^2 + 0.5 * 1*3.36^2 = 0.5 * 220 * A^2

A = 0.27 m

the amplitude of oscillation is 0.27 m

(b)

angular velocity , w = sqrt(K/m)

w = sqrt(220/1)

w = 14.83 rad/s

assume equation as

x = A * cos(w*t + phi)

therfore,

x = 0.27 * cos(14.83 *t + phi)

for t = 2.5 s

0.146 = 0.27 * cos(14.83*2.5 + phi)

phi = 1.671 rad

therefore, x = 0.27 * cos(14.83 *t + 1.671 )

at t= 0

x = 0.27 * cos(14.83 *0 + 1.671 )

x = - 0.027 m

(b)

the position at t=0 is - 0.027 m

(c)

at t = 0

velocity , v = x*w

v = 0.027 * 14.83

v = 0.401 m/s

the velocity at t = 0 is 0.401 m/s

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