A 200-g block is attached to a horizontal spring and executes simple harmonic motion with a period of 0.250 s. The total energy of the system is 2.00 J. Find (a) the force constant of the spring and (b) the amplitude of the motion.

Open in App

Solution

We are given $m = 200 g, T = 0.250 s, E = 2.00 J,$ and
$ω = \frac{2 π}{T} = \frac{2 π}{0.250} = 25.1 r a d / s$
(a) $k = m ω^{2} = (0.200 k g) (25.1 r a d / s)^{2} = 126 N / m$
(b) $E = \frac{k A^{2}}{2} ⟹ A = \sqrt{\frac{2 E}{k}} = \sqrt{\frac{2 (2.00 J)}{126 N / m}} = 0.178 m$

Suggest Corrections

Similar questions

Q. A 1 kg block is executing simple harmonic motion of amplitude 0.1 m on a smooth horizontal surface under the restoring force of a spring of spring constant 100 N m⁻¹. A block of mass 3 kg is gently placed on it at the instant it passes through the mean position. Assuming that the two blocks move together, find the frequency and the amplitude of the motion.

Figure

Q. derive the expression of displacement of a block which is attached to a spring constant k and executing simple harmonic motion

A block of mass 0.5 kg hanging from a vertical spring executes simple harmonic motion of amplitude 0.1 m and time period 0.314 s. Find the maximum force exerted by the spring on the block.

Q. A block of mass

2 k g

executes simple harmonic motion under the restoring force of a spring. The amplitude and the time period of motions are

0.2 c m and 2 π s e c

respectively. Find the maximum force exerted by the spring on the block.

A 1 kg block is executing simple harmonic motion of amplitude 0.1 m on a smooth horizontal surface under the restoring force of a spring of spring constant $100 N m^{- 1.}$ A block of mass 3 kg is gently placed on it at the instant it passes through the mean position. Assuming that the two blocks move together, find the frequency and the amplitude of the motion.

Join BYJU'S Learning Program

A 200-g block is attached to a horizontal spring and executes simple harmonic motion with a period of 0.250 s. The total energy of the system is 2.00 J. Find (a) the force constant of the spring and (b) the amplitude of the motion.

We are given m=200 g,T=0.250 s,E=2.00 J, and ω=2πT=2π0.250=25.1 rad/s(a) k=mω2=(0.200 kg)(25.1 rad/s)2=126 N/m(b) E=kA22⟹A=√2Ek=√2(2.00 J)126 N/m=0.178 m

We are given $m = 200 g, T = 0.250 s, E = 2.00 J,$ and
$ω = \frac{2 π}{T} = \frac{2 π}{0.250} = 25.1 r a d / s$
(a) $k = m ω^{2} = (0.200 k g) (25.1 r a d / s)^{2} = 126 N / m$
(b) $E = \frac{k A^{2}}{2} ⟹ A = \sqrt{\frac{2 E}{k}} = \sqrt{\frac{2 (2.00 J)}{126 N / m}} = 0.178 m$