A simple harmonic oscillator is of mass 0-100 kg- It is oscillating with a frequency of 5-Hz- If its amplitude of vibration is 5 cm- the force acting on the particle at its extreme position is

The correct option is D 0.5 N F #160; = #160; m #160; ω^2 #160; A #160; ω #160;= #160; 2π #160; f = 10 #160; #160; rad #160; / ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Forced Oscillations

A simple harm...

Question

A simple harmonic oscillator is of mass $0.100$ kg. It is oscillating with a frequency of $\frac{5}{π}$ Hz. If its amplitude of vibration is $5$ cm, the force acting on the particle at its extreme position is

2 N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.5 N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1 N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.5 N

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D 0.5 N
$F = m ω^{2} A$
$ω = 2 π f = 10 r a d / s e c$
$F = 0.1 \times 100 \times \frac{5}{100}$
$= 0.5 N$

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

. A block of mass

3 kg

is gently placed on it at the instant it passes through the mean position. Assuming that the blocks move together, then

Q. A particle is executing simple harmonic motion between extreme positions given by

(- 1. - 2. - 3) c m

and

(1, 2, 1) c m

. Its amplitude of oscillation is :

Q. A particle executing simple harmonic motion of amplitude 5 cm has maximum speed of 31.4 cm/s. The frequency of its oscillation is :

A simple harmonic oscillator of angular frequency $2$ rad/s is acted upon by an external force $F = sin t$ N. If the oscillator is at rest in its equilibrium position at $t = 0$ , its position at later times is proportional to:

Q. A simple harmonic oscillator of angular frequency

2 r a d s^{- 1}

is acted upon by an external force

F = sin t N

. If the oscillator is at rest in its equilibrium position at

t = 0

, its position at later times is proportional to

Join BYJU'S Learning Program

A simple harmonic oscillator is of mass 0.100 kg. It is oscillating with a frequency of 5πHz. If its amplitude of vibration is 5 cm, the force acting on the particle at its extreme position is

The correct option is D 0.5 NF=mω2A ω=2πf=10rad/secF=0.1×100×5100 =0.5N

A simple harmonic oscillator is of mass $0.100$ kg. It is oscillating with a frequency of $\frac{5}{π}$ Hz. If its amplitude of vibration is $5$ cm, the force acting on the particle at its extreme position is

The correct option is D 0.5 N
$F = m ω^{2} A$
$ω = 2 π f = 10 r a d / s e c$
$F = 0.1 \times 100 \times \frac{5}{100}$
$= 0.5 N$