0

Guest

Courses New

A 5.13-kg object moves on a horizontal frictionless surface under the influence of a spring with force constant 9.88 N/cm. The object is displaced 53.5 cm and given an initial velocity of 11.2 m/s back toward the equilibrium position. Find (a) the frequency of the motion, (b) he initial potential energy of the system, (c) the initial kinetic energy, and (d) the amplitude of the motion.

A 5.13-kg object moves on a horizontal frictionless surface under the influence of a spring with force constant 9.88 N/cm. The object is displaced 53.5 cm and given an initial velocity of 11.2 m/s back toward the equilibrium position. Find (a) the frequency of the motion, (b) he initial potential energy of the system, (c) the initial kinetic energy, and (d) the amplitude of the motion.

Grade:11

1 Answers

Jitender Pal
askIITians Faculty 365 Points
8 years ago
235-483_12.PNG
235-1251_13.PNG
= √(2×463 J)/(9.88 N/cm)
= √(2×463 J)/(9.88 N/cm×1 cm/10-2 m)
= 0.968 √J / N/m
= 0.968 √(J.m / N) (1 N.m/ 1J)
=0.968 m
From the above observation we conclude that, the amplitude xm of the motion would be 0.968 m.

Think You Can Provide A Better Answer ?

Other Related Questions on Wave Motion

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More