Question

A simple harmonic oscillator consists of a block attached to a spring, moving back and forth on a frictionless horizontal surface. Suppose the mass of the box is 5.0 kg. The motion is started by holding the box at 0.50 m from its central position, using a force of 40.0 N. Then the box is let go and allowed to perform simple harmonic motion.

"heereen + - 5 m m

(a) What is the amplitude of the motion?

(b) What is the spring constant k?

(c) What is the maximum elastic potential energy?

(d) What is the maximum speed of the block?

(e) At a point 0.20 m from the center of motion, what is the speed of the block?

(f) Where is the block when its speed is 1.0 m/s?

Answer 1

Solni 93 As TA= 40= N m L <C> Maxime num x=0.5 P.E. block is released from extreme position which is 0.5m from neutral point Se, Amplitude, A=0.5m Ang <b> In folding block at x=0.5 9 force required kx =) kr0.5 / k=80 Ang elastic potential energy will occur when - 2 kx2= ** * 80 260.572 you 7 Ane <d> speed of block will be maximum when it will pass from mean fosidson, Sor using energy balance, Energy at extreme = Energy at mean position KA² Ź my 2 2. = A /126 = d.om Angular fequency of oscillatori (P.E.) max 1 TA 2 = 0.5 * ४० 5 s kes 80 we F rad/see Loops =4 5 velocity of block when it is ad X20-2 m W SA²x² = + 0.5-0.22 m Vo 1.833 Ang <f> wheni vawlA2x2 3 v-1 m/s I=4 Jais ²x2 location when v=1m/s x = 0.433M J