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 m(b) calculate the maximum velocity attained by the object. [hint: use conservation of energy. m/s

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

here we know

m = 3.5 kg

k = 270 N/m

now we have

Part a)

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

Part b)

Maximum speed of SHM at its mean position is given as

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