Physics: Acart of mass 250 g is place on a frictionless horizontal air track. a spring having a spring constant of 9.5 n/m is attached between the cart and the left end o f the track. if the cart is displaced 4.5 cm from its equilibrium position, find: a) the period at which it oscillates.[1.0 s] b) its maximum speed.[0.28 m/s] c) its speed when it is located 2.0 cm from its equilibrium position.[0.25 m/s]

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Acart of mass 250 g is place on a frictionless

Ответ:

jasjas12

13.01.2022

Physics

a) T=1.02 s

b) $v=0.28\frac{m}{s}$

c) $v=0.25\frac{m}{s}$

Explanation:

a) The period is defined as:

$T=\frac{2\pi}{\omega}$

$\omega$ is the natural frequency of the system, in this case is given by:

$\omega=\sqrt{\frac{k}{m}}\\\omega=\sqrt{\frac{9.5\frac{N}{m}}{250*10^{-3}kg}}\\\omega=6.16\frac{rad}{s}$

Now, we calculate the period:

$T=\frac{2\pi}{6.16\frac{rad}{s}}\\T=1.02 s$

b) According to the law of conservation of energy, we have:

$\frac{kx^2}{2}=\frac{mv^2}{2}\\v=\sqrt{\frac{kx^2}{m}}\\v=\sqrt{\frac{(9.5\frac{N}{m})(4.5*10^{-2}m)^2}{250*10^{-3}kg}}\\v=0.28\frac{m}{s}$

c) In this case, we have:

$U_i=U_f+K_f\\\frac{kx_i^2}{2}=\frac{kx_f^2}{2}+\frac{mv^2}{2}\\v=\sqrt{\frac{k(x_i^2-x_f^2)}{m}}\\v=\sqrt{\frac{9.5\frac{N}{m}((4.5*10^{-2}m)^2-(2*10^{-2}m)^2)}{250*10^{-3}kg}}\\v=0.25\frac{m}{s}$

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