The total mechanical energy of a spring-mass | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The total mechanical energy of a spring-mass system is given by $E = \frac{1}{2}kA^2$,where $k$ is the spring constant and $A$ is the amplitude.We

Vedclass · Accepted Answer

remain same

Vedclass · Answer

(D) The total mechanical energy of a spring-mass system is given by $E = \frac{1}{2}kA^2$,where $k$ is the spring constant and $A$ is the amplitude.We know that for a spring-mass system,the angular frequency is $\omega = \sqrt{\frac{k}{m}}$,which implies $k = m\omega^2$.Substituting $k = m\omega^2$ into the energy equation,we get $E = \frac{1}{2}(m\omega^2)A^2$.When the mass $m$ is replaced by $2m$,the spring constant $k$ of the spring remains unchanged because the spring itself is not replaced.Since the total mechanical energy $E = \frac{1}{2}kA^2$ depends only on the spring constant $k$ and the amplitude $A$,and both $k$ and $A$ remain constant,the total mechanical energy $E$ remains the same.

The total mechanical energy of a spring-mass system in simple harmonic motion is $E = \frac{1}{2}m\omega^2 A^2$. Suppose the oscillating particle is replaced by another particle of double the mass while the amplitude $A$ remains the same. The new mechanical energy will:

Similar Questions

$A$ sphere of mass $m$ is attached to the lower end of a light vertical spring with force constant $k$. The upper end of the spring is fixed. The sphere is released from rest at the natural (unstretched) length of the spring and comes to rest again after falling through a distance $x$.

Springs of spring constants $K, 2K, 4K, 8K, \dots$ are connected in series. $A$ mass of $40 \, g$ is attached to the lower end of the last spring and the system is allowed to vibrate. What is the time period of oscillation in seconds? (Given $K = 2 \, N/cm$)

Three masses $500 \ g$,$300 \ g$,and $100 \ g$ are suspended at the end of a spring as shown in the figure and are in equilibrium. When the $500 \ g$ mass is removed,the system oscillates with a period of $2 \ s$. When the $300 \ g$ mass is also removed,it will oscillate with a period of (in $s$)

The frequency of oscillation of a mass $m$ suspended by a spring is $v_1$. If the length of the spring is cut to half, the same mass oscillates with frequency $v_2$. The value of $v_2/v_1$ is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module