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

Part a)

A = 0.066 m

Part b)

maximum speed = 0.58 m/s

Step-by-step explanation:

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

`\omega = \sqrt{(k)/(m)}`

here we know

m = 3.5 kg

k = 270 N/m

now we have

`\omega = \sqrt{(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 √(A^2 - x^2)`

`0.55 = 8.78 √(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`