A mass hangs from a spring and oscillates ver | vedclass

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

Question

(B) Let $M$ be the mass of the box and $m$ be the mass of the oscillating object. The scale measures the normal force $N$ exerted by the box on the sc

Vedclass · Accepted Answer

At its minimum height

Vedclass · Answer

(B) Let $M$ be the mass of the box and $m$ be the mass of the oscillating object. The scale measures the normal force $N$ exerted by the box on the scale. For the whole system (box + spring + mass),the forces acting are the weight $(M+m)g$ downwards and the normal force $N$ upwards. From Newton's second law for the mass $m$,the force exerted by the spring on the mass is $F_s = m(g + a)$,where $a$ is the acceleration of the mass (upward is positive). By Newton's third law,the mass exerts an equal and opposite force $F_s$ on the spring,which is transmitted to the box. Thus,the total normal force $N$ on the scale is $N = Mg + F_s = Mg + m(g + a) = (M+m)g + ma$. The scale reading is largest when the acceleration $a$ is at its maximum positive value (upward). In simple harmonic motion,the acceleration is directed towards the equilibrium position. At the minimum height (lowest point),the restoring force is upward and maximum,so the acceleration $a$ is at its maximum positive value. Therefore,the scale reading is largest when the mass is at its minimum height.

$A$ mass hangs from a spring and oscillates vertically. The top end of the spring is attached to the top of a box,and the box is placed on a scale,as shown in the figure. The reading on the scale is largest when the mass is

Similar Questions

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:

$A$ simple harmonic oscillator consists of a particle of mass $m$ and an ideal spring with spring constant $k$. The particle oscillates with a time period $T$. The spring is cut into two equal parts. If one part oscillates with the same particle, the new time period will be

Let $T_1$ and $T_2$ be the time periods of two springs $A$ and $B$ when a mass $m$ is suspended from them separately. Now both the springs are connected in parallel and the same mass $m$ is suspended with them. If $T$ is the new time period in this position,then:

$A$ force of $6.4 \, N$ stretches a vertical spring by $0.1 \, m$. The mass that must be suspended from the spring so that it oscillates with a period of $\left( \frac{\pi}{4} \right) \, s$ is ... $kg$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module