A mass is suspended from a vertical spring wh | vedclass

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

Question

(D) The frequency of $S.H.M.$ is given by $n = 5 Hz$. The angular frequency is $\omega = 2 \pi n = 2 \pi 	imes 5 = 10 \pi rad s^{-1}$.At the highest

Vedclass · Accepted Answer

$\frac{1}{\pi} m s^{-1}$

Vedclass · Answer

(D) The frequency of $S.H.M.$ is given by $n = 5 Hz$. The angular frequency is $\omega = 2 \pi n = 2 \pi 	imes 5 = 10 \pi rad s^{-1}$.At the highest point of oscillation,the spring is unstretched,meaning the extension $x = 0$. In $S.H.M.$,the equilibrium position is where the spring force balances gravity,$k x_0 = mg$,where $x_0$ is the static extension.The amplitude $A$ of the oscillation is equal to this static extension $x_0$,because the particle reaches the unstretched position $(x=0)$ at the top of its path. Thus,$A = x_0 = \frac{mg}{k}$.We know that $\omega^2 = \frac{k}{m}$,so $k = m \omega^2 = m(10 \pi)^2 = 100 \pi^2 m$.Substituting $k$ into the amplitude equation: $A = \frac{mg}{100 \pi^2 m} = \frac{g}{100 \pi^2} = \frac{10}{100 \pi^2} = \frac{1}{10 \pi} m$.The maximum speed is $V_{max} = \omega A = (10 \pi) 	imes (\frac{1}{10 \pi}) = 1 m s^{-1}$.Wait,re-evaluating the calculation: $V_{max} = \omega A = (10 \pi) 	imes \frac{g}{\omega^2} = \frac{g}{\omega} = \frac{10}{10 \pi} = \frac{1}{\pi} m s^{-1}$.

$A$ mass is suspended from a vertical spring which is executing $S.H.M.$ of frequency $5 Hz$. The spring is unstretched at the highest point of oscillation. The maximum speed of the mass is [acceleration due to gravity $g=10 m s^{-2}$].

Similar Questions

$A$ mass of $0.2\,kg$ is attached to the lower end of a massless spring of force-constant $200\,N/m,$ the upper end of which is fixed to a rigid support. Which of the following statements is/are true?

Two masses $m_1$ and $m_2$ are suspended together by a massless spring of constant $K$. When the masses are in equilibrium,$m_1$ is removed without disturbing the system. The amplitude of oscillations is

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 $K = 2 \, N/cm$)

$A$ 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

Two springs of force constants $2k$ and $k$ are connected to a mass $m$ as shown. The mass is displaced slightly to one side and released. The frequency of oscillation of the two springs-mass system is

Vedclass Test Series

Exam Paper Generator

Online Exam Module