a. A block with a mass of 0.352 kg is attached to a spring of spring constant 57

Question

a. A block with a mass of 0.352 kg is attached to a spring of spring constant 572 N/m. It is sitting at equilibrium. You then pull the block down 3.30 cm from equilibrium and let go. What is the amplitude of the oscillation?

b. A block with a mass of 0.704 kg is attached to a spring of spring constant 572 N/m. It is sitting at equilibrium. You then pull the block down 3.30 cm from equilibrium and let go. What is the amplitude of the oscillation?

c. A block with a mass of 0.352 kg is attached to a spring of spring constant 1.14×103 N/m. It is sitting at equilibrium. You then pull the block down 3.30 cm from equilibrium and let go. What is the amplitude of the oscillation?

d. A block with a mass of 0.352 kg is attached to a spring of spring constant 572 N/m. It is sitting at equilibrium. You then pull the block down 6.60 cm from equilibrium and let go. What is the amplitude of the oscillation?

Explanation / Answer

(a) Given data is

What is the amplitude of the oscillation?

Solution::--

here mass m= 0.352 kg

spring constant K= 572 N/m.

x=3.30 cm=0.033 m

Now we find the angular frequency w = sqrt(K/m)= sqrt(572/0.352) = sqrt (1625) = 40.315 rad/sec

Where as angular frequency w = 2f = 2 / t

frequency f =w/2 = 40.315/(2*3.14) = 40.315/6.28 = 6.419 Hz

from frequency we can findout the value of time period per seconds t=1/f = 1/6.419 = 0.155 sec

amplitude A = x/[cos(wt)] = 0.033/(cos(40.315*0.155)) = 0.033/0.99 = 0.033m

The amplitude of the oscillation (A) = 0.033 m

(b) given data is

What is the amplitude of the oscillation?

Solution::--

here mass m= 0.704 kg

spring constant K= 572 N/m.

x=3.30 cm=0.033 m

Now we find the angular frequency w = sqrt(K/m)= sqrt(572/0.704) = sqrt (812.5) = 28.50 rad/sec

Where as angular frequency w = 2f = 2 / t

frequency f =w/2 = 28.50/(2*3.14) = 28.50/6.28 = 4.53 Hz

from frequency we can findout the value of time period per seconds t=1/f = 1/4.53 = 0.220 sec

amplitude A = x/[cos(wt)] = 0.033/(cos(28.50*0.220)) = 0.033/0.994 = 0.0331 m

The amplitude of the oscillation (A) = 0.0331 m

(c) Given data is

What is the amplitude of the oscillation?

Solution::--

here mass m= 0.352 kg

spring constant K= 1.14*103 N/m.=1140 N/m

x=3.30 cm=0.033 m

Now we find the angular frequency w = sqrt(K/m)= sqrt(1140/0.352) = sqrt (3238.6) = 56.90 rad/sec

Where as angular frequency w = 2f = 2 / t

frequency f =w/2 = 56.90/(2*3.14) = 56.90/6.28 = 9.06 Hz

from frequency we can findout the value of time period per seconds t=1/f = 1/9.06 = 0.11 sec

amplitude A = x/[cos(wt)] = 0.033/(cos(56.90*0.11)) = 0.033/0.99 = 0.033m

The amplitude of the oscillation (A) = 0.033 m

(d) Given data is

What is the amplitude of the oscillation?

Solution::--

here mass m= 0.352 kg

spring constant K= 572 N/m.

x=6.60 cm=0.066 m

Now we find the angular frequency w = sqrt(K/m)= sqrt(572/0.352) = sqrt (1625) = 40.315 rad/sec

Where as angular frequency w = 2f = 2 / t

frequency f =w/2 = 40.315/(2*3.14) = 40.315/6.28 = 6.419 Hz

from frequency we can findout the value of time period per seconds t=1/f = 1/6.419 = 0.155 sec

amplitude A = x/[cos(wt)] = 0.066/(cos(40.315*0.155)) = 0.033/0.99 = 0.066m

The amplitude of the oscillation (A) = 0.066 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.