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jek_recluse [69]
3 years ago
13

A block is attached to a horizontal spring and oscillates back and forth on a frictionless horizontal surface at a frequency of

4.42 Hz. The amplitude of the motion is 8.26 x 10-2 m. At the point where the block has its maximum speed, it suddenly splits into two identical parts, only one part remaining attached to the spring.
(a) What is the amplitude and
(b) the frequency of the simple harmonic motion that exists after the block splits?
Physics
1 answer:
Nastasia [14]3 years ago
8 0

Answer:

Amplitude = 0.058m

Frequency = 6.25Hz

Explanation:

Given

Amplitude (A) = 8.26 x 10-2 m

Frequency (f) = 4.42Hz

Conversation of energy before split

½mv² = ½KA²

Make A the subject of formula

A = v\sqrt{\frac{m}{k} }

Conversation of energy after split

½(m/2)V'² = ½(m/2)V² = ½KA'²

½(m/2)V² = ½KA'²

Make A the subject of formula

First divide both sides by ½

(m/2)V² = KA'²

Divide both sides by K

\frac{m}{2K}V² = A'²

v\sqrt{\frac{m}{2k} } = A'

Substitute v\sqrt{\frac{m}{k} } for A in the above equation

A' = A/√2

A' = 8.26 x 10^-2/√2

A' = 0.05840702012600882

Amplitude after split = 0.058 (Approximated)

Frequency (f') = f√2

f' = 4.42√2

f' = 6.25082394568908011

Frequency after split = 6.25Hz (approximated)

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2 years ago
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Answer:

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Explanation:

Part A

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v = 14.4 m/s

acceleration at the top is given as

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now we know the formula of centripetal acceleration as

a = \frac{v^2}{R}\\

19.6 = \frac{14.4^2}{R}\\R = 10.58 m

Part B

now

the total mass of the car and the ride is M

Let the force exerted by the track be n

By Newton law

n +Mg =\frac{Mv^2}{r} \\\\n=\frac{Mv^2}{r} -Mg\\\\=M(\frac{v^2}{r}-g )\\\\=M(2g-g)\\\\T=Mg

Part C

If the radius of the loop is 21.4 m

speed is given by same v = 14.4 m/s

now the acceleration is given as

a = \frac{v^2}{R}

a = \frac{14.4^2}{21.4} \\\\= 9.69 m/s^2

Now for normal force at the top is given by force equation

F_n + mg = ma\\F_n = m(a-g)

The force exerted by the rail is less than zero because acceleration is less than 9.69m/s²

So the normal force would have to point away from the centre, For safe ride this normal force must be positive i.e a \prec g

\frac{v^2}{r} \prec \sqrt{g} \\\\v = \sqrt{rg} \\\\v = \sqrt{21.4 \times 9.8} \\\\v = 14.5m/s

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