Question

A cart 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]

Answer 1

Answer:

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:

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

Answer 2

The maximum speed is; 0.28 m/s

The speed when it is located 2.0 cm from its equilibrium position is; 0.25 m/s

What is the speed from the equilibrium position?

A) Formula for period is;

T = √(2π/ω)

where ω which is the natural angular frequency

ω = √(k/m)

we are given;

k = 9.5 N/m

m = 250 g = 0.25 kg

x = 4.5 cm = 0.045 m

Thus;

ω = √(9.5/0.25)

ω = 6.1 rad/s

Thus;

Period is; T = √(2π/6.1)

T = 1.02 s

B) From law of conservation of energy;

maximum speed is;

v = √(kx²/m)

v = √(9.5 * 0.045²/0.25)

v = 0.28 m/s

C) We are now given;

x₁ = 4.5cm = 0.045 m

x₂ = 2 cm = 0.02 m

Thus, speed is gotten from;

v = √((k(x₁ - x₂)/m)

v = √((9.5(0.045 - 0.02)/0.25)

v = 0.25 m/s

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