A 10 kg collar is attached to a spring (spring constant 600 N/m), it slides without friction over a horizontal rod. The collar is displaced from its equilibrium position by 20 cm and released. What is the speed of the oscillation?

Question

A 10 kg collar is attached to a spring (spring constant 600 N/m), it slides without friction over a horizontal rod. The collar is displaced from its equilibrium position by 20 cm and released. What is the speed of the oscillation? (A)  √60 × 0.2 m/s (B)  60 × 0.2 m/s (C)  60 × 2 m/s (D)  6 × 0.2 m/s

Tardigrade Admin · Accepted Answer

Angular frequency of spring - block system is given by

ω = \frac{k}{m}

Maximum speed in oscillation

v_{m a x} = A ω = \frac{k}{m}

Here,

A = 20 c m = 0.2 m, k = 600 N m^{- 1}

m = 10 k g, v_{m a x} = ?

v_{m a x} = 0.2 \frac{600}{10} = 0.2 60 m s^{- 1}

Q. A $10 k g$ collar is attached to a spring (spring constant $600 N / m$ ), it slides without friction over a horizontal rod. The collar is displaced from its equilibrium position by $20 c m$ and released. What is the speed of the oscillation?

$60 \times 0.2 m / s$

$60 \times 0.2 m / s$

$60 \times 2 m / s$

$6 \times 0.2 m / s$

Angular frequency of spring - block system is given by
$ω = \frac{k}{m}$
Maximum speed in oscillation
$v_{m a x} = A ω = \frac{k}{m}$
Here, $A = 20 c m = 0.2 m, k = 600 N m^{- 1}$
$m = 10 k g, v_{m a x} = ?$
$v_{m a x} = 0.2 \frac{600}{10} = 0.2 60 m s^{- 1}$

Q. A 10kg collar is attached to a spring (spring constant 600N/m), it slides without friction over a horizontal rod. The collar is displaced from its equilibrium position by 20cm and released. What is the speed of the oscillation?

60​×0.2m/s

60×0.2m/s

60×2m/s

6×0.2m/s

Angular frequency of spring - block system is given by ω=mk​​ Maximum speed in oscillation vmax​=Aω=mk​​ Here, A=20cm=0.2m,k=600Nm−1 m=10kg,vmax​=? vmax​=0.210600​​=0.260​ms−1

Q. A $10 k g$ collar is attached to a spring (spring constant $600 N / m$ ), it slides without friction over a horizontal rod. The collar is displaced from its equilibrium position by $20 c m$ and released. What is the speed of the oscillation?

$60 \times 0.2 m / s$

$60 \times 0.2 m / s$

$60 \times 2 m / s$

$6 \times 0.2 m / s$

Angular frequency of spring - block system is given by
$ω = \frac{k}{m}$
Maximum speed in oscillation
$v_{m a x} = A ω = \frac{k}{m}$
Here, $A = 20 c m = 0.2 m, k = 600 N m^{- 1}$
$m = 10 k g, v_{m a x} = ?$
$v_{m a x} = 0.2 \frac{600}{10} = 0.2 60 m s^{- 1}$