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Butoxors [25]
3 years ago
11

A force of 960 newtons stretches a spring 4 meters. A mass of 60 kilograms is attached to the end of the spring and is initially

released from the equilibrium position with an upward velocity of 6 m/s. Find the equation of motion.
Physics
1 answer:
Drupady [299]3 years ago
4 0

Answer:

x(t) = - 6 cos 2t

Explanation:

Force of spring = - kx

k= spring constant

x= distance traveled by compressing

But force = mass × acceleration

==> Force = m × d²x/dt²

===> md²x/dt² = -kx

==> md²x/dt² + kx=0   ------------------------(1)

Now Again, by Hook's law

Force = -kx

==> 960=-k × 400

==> -k =960 /4 =240 N/m

ignoring -ve sign k= 240 N/m

Put given data in eq (1)

We get

60d²x/dt² + 240x=0

==> d²x/dt² + 4x=0

General solution for this differential eq is;

x(t) = A cos 2t + B sin 2t   ------------------------(2)

Now initially

position of mass spring

at time = 0 sec

x (0) = 0 m

initial velocity v= = dx/dt=  6m/s

from (2) we have;

dx/dt= -2Asin 2t +2B cost 2t = v(t) --- (3)

put t =0 and dx/dt = v(0) = -6 we get;

-2A sin 2(0)+2Bcos(0) =-6

==> 2B = -6

B= -3

Putting B = 3 in eq (2) and ignoring first term (because it is not possible to find value of A with given initial conditions) - we get

x(t) = - 6 cos 2t

==>  

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For a body to be in SHM, the restoring force must be __________ to displacement from equilibrium.
Ipatiy [6.2K]

Answer:

Proportional

Explanation:

The conditions that must be met to produce SHM are;

-The restoring force must be proportional to the displacement and act opposite to the direction of motion with no drag forces or friction.

- The frequency of oscillation does not depend on the amplitude.

3 0
2 years ago
During an experiment, you take a measurement of 12.9 inches.
svlad2 [7]

Answer:

1 inch = 2.54 cm

12.9 inches= 12.9 x 2.54

= 32.766

= 32.8 cm (approximately)

Hope it helps...

7 0
3 years ago
If you had started with a larger mass, how would the half-life change?
iogann1982 [59]

Answer:

There is no change, unless your mass is somehow at the quantum level, at which the concept of half-life breaks down.

Half life is a property of the specific radioactive isotope...NOT of the initial sample's mass.

3 0
3 years ago
A body of mass 2.7 kg makes an elastic collision with another body at rest and continues to move in the original direction but w
kramer

Answer:

a)

1.35 kg

b)

2.67 ms⁻¹

Explanation:

a)

m_{1} = mass of first body = 2.7 kg

m_{2} = mass of second body = ?

v_{1i} = initial velocity of the first body before collision = v

v_{2i} = initial velocity of the second body before collision = 0 m/s

v_{1f} = final velocity of the first body after collision =

using conservation of momentum equation

m_{1} v_{1i} + m_{2} v_{2i} = m_{1} v_{1f} + m_{2} v_{2f}\\(2.7) v + m_{2} (0) = (2.7) (\frac{v}{3} ) + m_{2} v_{2f}\\(2.7) (\frac{2v}{3} ) = m_{2} v_{2f}\\v_{2f} = \frac{1.8v}{m_{2}}

Using conservation of kinetic energy

m_{1} v_{1i}^{2}+ m_{2} v_{2i}^{2} = m_{1} v_{1f}^{2} + m_{2} v_{2f}^{2} \\(2.7) v^{2} + m_{2} (0)^{2} = (2.7) (\frac{v}{3} )^{2} + m_{2} (\frac{1.8v}{m_{2}})^{2} \\(2.7) = (0.3) + \frac{3.24}{m_{2}}\\m_{2} = 1.35

b)

m_{1} = mass of first body = 2.7 kg

m_{2} = mass of second body = 1.35 kg

v_{1i} = initial velocity of the first body before collision = 4 ms⁻¹

v_{2i} = initial velocity of the second body before collision = 0 m/s

Speed of the center of mass of two-body system is given as

v_{cm} = \frac{(m_{1} v_{1i} + m_{2} v_{2i})}{(m_{1} + m_{2})}\\v_{cm} = \frac{((2.7) (4) + (1.35) (0))}{(2.7 + 1.35)}\\\\v_{cm} = 2.67 ms⁻¹

8 0
3 years ago
If the distance between slits on a diffraction grating is 0.50 mm and one of the angles of diffraction is 0.25°, how large is th
Trava [24]

Answer:

-  path differnce = 2.18*10^-6

-  1538 lines

Explanation:

- The path difference for the waves that produce the pattern of diffraction, is given by the following formula:

path\ difference\ = dsin\theta           (1)

d: separation between slits = 0.50mm = 0.50*10^-3 m

θ: angle of a diffraction = 0.25°

Then, the path difference is:

path\ difference\ =(0.50*10^{-3}m)sin(0.25\°)=2.18*10^{-6}m

- The maximum number of bright lines are calculated by using the following formula:

m\lambda = dsin\theta           (2)

m: order of the bright

λ: wavelength = 650nm

The maximum bright is calculated for an angle of 90°:

m=\frac{(0.50*10^{-3}m)sin90\°}{650*10^{-9}m} \approx 769

The maxium number of bright lines are twice the previous result, that is, 1538 lines

8 0
3 years ago
Read 2 more answers
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