Question

An object with mass 3.5 kg is attached to a spring with spring stiffness constant k = 270 N/m and is executing simple harmonic motion. When the object is 0.020 m from its equilibrium position, it is moving with a speed of 0.55 m/s.(a) Calculate the amplitude of the motion._____ m(b) Calculate the maximum velocity attained by the object. [Hint: Use conservation of energy.]______ m/s

Answer 1

Answer:

Part a)

A = 0.066 m

Part b)

maximum speed = 0.58 m/s

Explanation:

As we know that angular frequency of spring block system is given as

$\omega = \sqrt{\frac{k}{m}}$

here we know

m = 3.5 kg

k = 270 N/m

now we have

$\omega = \sqrt{\frac{270}{3.5}}$

$\omega = 8.78 rad/s$

Part a)

Speed of SHM at distance x = 0.020 m from its equilibrium position is given as

$v = \omega \sqrt{A^2 - x^2}$

$0.55 = 8.78 \sqrt{A^2 - 0.020^2}$

$A = 0.066 m$

Part b)

Maximum speed of SHM at its mean position is given as

$v_{max} = A\omega$

$v_{max} = 0.066(8.78) = 0.58 m/s$