SATHEE: Simple Harmonic Motion 1 Question 6

Simple Harmonic Motion 1 Question 6

6. A linear harmonic oscillator of force constant $2 \times 10^{6} N / m$ and amplitude $0.01 m$ has a total mechanical energy of $160 J$ . Its

$(1989, 2 M)$

(a) maximum potential energy is $100 J$

(b) maximum kinetic energy is $100 J$

(d) maximum potential energy is zero

Fill in the Blank

Show Answer

Answer:

Correct Answer: 6. $(b, c)$

Solution:

$\frac{1}{2} k A^{2} = \frac{1}{2} \times 2 \times 10^{6} \times {(10^{- 2})}^{2} = 100 J$

This is basically the energy of oscillation of the particle.

$K, U$ and $E$ at mean position $(x = 0)$ and extreme position $(x = \pm A)$ are shown in figure.

$\begin{array}{ll} x = 0 \\ K = 100 J = Maximum & K = 0 J \\ U = 60 J = Minimum & U = 160 J = Maximum \\ E = 160 J = Constant & E = 160 J = Constant \end{array}$

$∴$ Correct options are (b) and (c).

Simple Harmonic Motion 1 Question 6

6. A linear harmonic oscillator of force constant $2 \times 10^{6} N / m$ and amplitude $0.01 m$ has a total mechanical energy of $160 J$ . Its

Fill in the Blank

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Simple Harmonic Motion 1 Question 6

6. A linear harmonic oscillator of force constant 2×106N/m and amplitude 0.01m has a total mechanical energy of 160J. Its

Fill in the Blank

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu

6. A linear harmonic oscillator of force constant $2 \times 10^{6} N / m$ and amplitude $0.01 m$ has a total mechanical energy of $160 J$ . Its