Answer:
Explanation:
Given that,
Mass attached m = 0.95kg
Spring constant k = 16N/m
Instantaneous speed v = 36cm/s = 0.36m/s
Amplitude A=?
When x = 0.7A
Using conservation of energy
∆K.E + ∆P.E = 0
K.E(final) — K.E(initial) + P.E(final) — P.E(initial) = 0
At the beginning immediately the hammer hits the mass, the potential energy is 0J, Therefore, P.E(initial) = 0J, so the speed is maximum.
Also, at the end, at maximum displacement, the speed is zero, therefore, K.E(final) = 0
So, the equation becomes
— K.E(initial) + P.E(final) = 0
K.E(initial) = P.E(final)
½mv² = ½kA²
mv² = kA²
0.95 × 0.36² = 16×A²
0.12312 = 16•A²
A² = 0.12312/16
A² = 0.007695
A = √0.007695
A = 0.088 m
A = 8.8cm
B. Speed at x = 0.7A
Using the same principle above
K.E(initial) = P.E(final)
½mv² = ½kA²
Where A = 0.7A = 0.7 × 0.088 = 0.0614m
Then,
½× 0.95 × v² = ½ × 16 × 0.0614²
0.475v² = 0.0310644
v² = 0.0310644/0.475
v² = 0.0635
v = √0.0635
v = 0.252 m/s
v = 25.2 cm/s
