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A 0.150-kg toy is undergoing SHM on the end of a horizontal spring with force constant k = 300 N/m. When the toy is 0.0120 m from its equilibrium position, it is observed to have a speed of 0.400 m/s. What are the toy鈥檚 (a) total energy at any point of its motion; (b) amplitude of motion; (c) maximum speed during its motion?

Short Answer

Expert verified
  1. Total energy at any point of its motion is E = 0.034 J.
  2. The amplitude of motion A = 1.5 cm.
  3. The maximum speed during its motion is Vmax=0.67 m/s.

Step by step solution

01

Calculate mechanical energy in SHM

a)

We know that,

E=12mvx2+12Kx2=12KA2 鈥︹︹︹︹ (1)

Here m is the mass of the object,is the velocity, K is the force constant of restoring force and A is the amplitude SHM.

E=120.15Kg(0.4m/s)2+12(300N/m)(0.012)2=0.034J

02

Calculate the amplitude of SHM

From equation (1)

A=2EK 鈥︹︹︹︹ (2)

role="math" localid="1668097274023" =20.034J300N/m=1.5cm

03

Calculate the maximum speed of SHM

Maximum velocity is at the equilibrium position, hence total energy is kinetic energy, and potential energy is 0. Hence, at equilibrium position

E=12mvmax2 鈥︹︹︹︹ (3)

From equation (3) is,

vmax=2Em=20.034J0.15Kg=0.67m/s

Hence, total energy of SHM is 0.034 J, amplitude of SHM is 1.5 cm and maximum velocity during the motion is 0.67 m/s.

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