Question

Груз массой 2 кг, закреплённый на пружине жёсткостью 200 Н/м, совершает гармонические колебания . Максимальная скорость груза при этом равна 1 м/с. Какова амплитуда колебаний груза?

Answer 1

$v_m=\omega x_m$

Следовательно, 

$x_m=\frac{v_m}{\omega}$

$v_m=1$ м/с

$\omega=\sqrt {\frac km}=10 \frac 1c$

 

$x_m=0.1$м

Answer 2

Calculation of Amplitude of Oscillations

To calculate the amplitude of the oscillations of the mass attached to the spring, we can use the equation for simple harmonic motion:

A = v_max / ω

Where: - A is the amplitude of the oscillations - v_max is the maximum velocity of the mass (given as 1 m/s) - ω is the angular frequency of the oscillations

To find the angular frequency, we can use the equation:

ω = √(k / m)

Where: - k is the spring constant (given as 200 N/m) - m is the mass of the object (given as 2 kg)

Let's calculate the amplitude using the given values:

ω = √(200 N/m / 2 kg) = √(100) = 10 rad/s

A = 1 m/s / 10 rad/s = 0.1 m

Therefore, the amplitude of the oscillations of the mass is 0.1 meters.