GRADE 9 PRACTICE
Physics Problems: Mechanical oscillations, Grade 9
Pendulums, harmonic, free and forced vibrations.
Here the conditions and analysis of the solution can be read without an account. In the simulator, you need to collect the same analyzes yourself: “Given,” formulas, order of steps and answer.
Problem statements
The answer options are there to check against, and the walkthrough opens under each problem.
Problem #28
Oscillation parameters and pendulumsWhat concept does the description correspond to: “The system is capable of repeatedly moving around a stable equilibrium position due to the restoring effect and inertia.”?
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- \text{Description}
- \text{A system can move repeatedly about stable equilibrium through restoring action and inertia.}
\text{name concepts} - ?
- \text{Let's compare approval with definition}
In both a pendulum and a mass–spring system, energy shifts between kinetic and potential forms.
- \text{Let's formulate conclusion}\text{Concept} = \text{Oscillatory system}
Answer\text{Concept} = \text{Oscillatory system}
Problem #29
Oscillation parameters and pendulumsWhat concept does the description correspond to: “Amplitude specifies the maximum deviation, period specifies the cycle time, and frequency specifies the number of cycles per second.”?
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- \text{Description}
- \text{Amplitude gives the maximum displacement; period gives cycle time; frequency gives cycles per second.}
\text{name concepts} - ?
- \text{Let's compare approval with definition}
Period and frequency are reciprocals; measurements over several cycles reduce relative error.
- \text{Let's formulate conclusion}\text{Concept} = \text{Amplitude, period and frequency}
Answer\text{Concept} = \text{Amplitude, period and frequency}
Problem #30
OscillationsThe pendulum completed 20 complete oscillations in 10 s. Find the period and frequency.
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- N
- 20
- t
- 10\,\text{с}
T - ?\nu - ?
- T=\frac{t}{N}T=\frac{10}{20}\,\text{с}T = 0{,}5\,\text{с}
- \nu=\frac{N}{t}\nu=\frac{20}{10}\,\text{Гц}\nu = 2\,\text{Гц}
Answer0{,}5\,\text{с}2\,\text{Гц}
Problem #31
Harmonic, damped and forced oscillationsWhat concept does the description correspond to: “Damped oscillations lose amplitude due to resistance, while forced oscillations are supported by an external periodic force.”?
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- \text{Description}
- \text{Damped oscillations decrease in amplitude because of resistance; forced oscillations are maintained by an external periodic force.}
\text{name concepts} - ?
- \text{Let's compare approval with definition}
After the transient dies away, the driven system oscillates at the driving frequency.
- \text{Let's formulate conclusion}\text{Concept} = \text{Damped and forced oscillations}
Answer\text{Concept} = \text{Damped and forced oscillations}
Problem #32
Harmonic, damped and forced oscillationsWhat concept does the description correspond to: “Oscillations in which the coordinate changes according to the law of sine or cosine with a constant frequency.”?
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- \text{Description}
- \text{Oscillations whose coordinate follows a sine or cosine law at constant frequency.}
\text{name concepts} - ?
- \text{Let's compare approval with definition}
Phase specifies the oscillator's state at a chosen instant and allows different oscillations to be compared.
- \text{Let's formulate conclusion}\text{Concept} = \text{Harmonic oscillations}
Answer\text{Concept} = \text{Harmonic oscillations}
Problem #33
OscillationsThe natural frequency of a weakly damped system is 5 Hz. At what frequency of a weak periodic force is the largest steady-state amplitude expected?
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- \nu_0
- 5\,\text{Гц}
\nu - ?
- \nu\approx\nu_0
Resonance occurs when the frequency of influence is close to the natural frequency.
- \nu=5\,\text{Гц}
Answer5\,\text{Гц}
Problem #118
Oscillation parameters and pendulumsFor 30 with the pendulum committed 20 complete hesitation. Find the period.
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- t
- 30\,\text{с}
- N
- 20\,\text{колебаний}
T - ?
- T=\frac{t}{N}T=\frac{30}{20}T=1{,}5\,\text{с}
A period is the time of one complete oscillation.
Answer1{,}5\,\text{с}
Problem #119
Oscillation parameters and pendulumsFor 40 with the pendulum committed 25 complete hesitation. Find the period.
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- t
- 40\,\text{с}
- N
- 25\,\text{колебаний}
T - ?
- T=\frac{t}{N}T=\frac{40}{25}T=1{,}6\,\text{с}
A period is the time of one complete oscillation.
Answer1{,}6\,\text{с}
Problem #120
Oscillation parameters and pendulumsOscillation period 0{,}5 s. How many complete oscillations will occur in 2 min?
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- T
- 0{,}5\,\text{с}
- t
- 2\,\text{мин}
N - ?
- a=60ta=2 \cdot 60a=120\,\text{с}
We reduce time to seconds.
- N=\frac{a}{T}N=\frac{120}{0{,}5}N=240\,\text{колебаний}
We divide the total time into a period.
Answer240\,\text{колебаний}
Problem #121
Oscillation parameters and pendulumsOscillation period 0{,}8 s. How many complete oscillations will occur in 4 min?
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- T
- 0{,}8\,\text{с}
- t
- 4\,\text{мин}
N - ?
- a=60ta=4 \cdot 60a=240\,\text{с}
We reduce time to seconds.
- N=\frac{a}{T}N=\frac{240}{0{,}8}N=300\,\text{колебаний}
We divide the total time into a period.
Answer300\,\text{колебаний}
Problem #122
Oscillation parameters and pendulumsOscillation amplitude 5 cm, period 2 s. How far in meters will the point travel? 20 s? Time is equal to an integer number of periods.
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- A
- 5\,\text{см}
- T
- 2\,\text{с}
- t
- 20\,\text{с}
S - ?
- N=\frac{t}{T}N=\frac{20}{2}N=10\,\text{колебаний}
We determine the number of complete oscillations.
- a=\frac{A}{100}a=\frac{5}{100}a=0{,}05\,\text{м}
Convert the amplitude to meters.
- S=4aNS=4 \cdot 0{,}05 \cdot 10S=2\,\text{м}
During a period, a point passes through four amplitudes.
Answer2\,\text{м}
Problem #123
Oscillation parameters and pendulumsOscillation amplitude 8 cm, period 4 s. How far in meters will the point travel? 40 s? Time is equal to an integer number of periods.
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- A
- 8\,\text{см}
- T
- 4\,\text{с}
- t
- 40\,\text{с}
S - ?
- N=\frac{t}{T}N=\frac{40}{4}N=10\,\text{колебаний}
We determine the number of complete oscillations.
- a=\frac{A}{100}a=\frac{8}{100}a=0{,}08\,\text{м}
Convert the amplitude to meters.
- S=4aNS=4 \cdot 0{,}08 \cdot 10S=3{,}2\,\text{м}
During a period, a point passes through four amplitudes.
Answer3{,}2\,\text{м}
Problem #124
Harmonic, damped and forced oscillationsPeriod of free oscillations of the system 0{,}25 s. Find your natural frequency.
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- T
- 0{,}25\,\text{с}
f - ?
- f=\frac{1}{T}f=\frac{1}{0{,}25}f=4\,\text{Гц}
Frequency and period are mutually inverse.
Answer4\,\text{Гц}
Problem #125
Harmonic, damped and forced oscillationsPeriod of free oscillations of the system 0{,}2 s. Find your natural frequency.
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- T
- 0{,}2\,\text{с}
f - ?
- f=\frac{1}{T}f=\frac{1}{0{,}2}f=5\,\text{Гц}
Frequency and period are mutually inverse.
Answer5\,\text{Гц}
Problem #126
Harmonic, damped and forced oscillationsFor 12 with the system committed 30 free vibrations. What frequency of external influence should be chosen for resonance with weak attenuation?
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- t
- 12\,\text{с}
- N
- 30\,\text{колебаний}
f - ?
- T=\frac{t}{N}T=\frac{12}{30}T=0{,}4\,\text{с}
Finding our own period.
- f=\frac{1}{T}f=\frac{1}{0{,}4}f=2{,}5\,\text{Гц}
With weak damping, the resonance is close to its natural frequency.
Answer2{,}5\,\text{Гц}
Problem #127
Harmonic, damped and forced oscillationsFor 20 with the system committed 60 free vibrations. What frequency of external influence should be chosen for resonance with weak attenuation?
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- t
- 20\,\text{с}
- N
- 60\,\text{колебаний}
f - ?
- T=\frac{t}{N}T=\frac{20}{60}T=0{,}333333\,\text{с}
Finding our own period.
- f=\frac{1}{T}f=\frac{1}{0{,}333333}f=3\,\text{Гц}
With weak damping, the resonance is close to its natural frequency.
Answer3\,\text{Гц}
Problem #128
Harmonic, damped and forced oscillationsInitial amplitude 20 cm. After each complete oscillation it is 0{,}5 from the previous one. Find the amplitude after three vibrations.
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- A
- 20\,\text{см}
- r
- 0{,}5\,\text{доля}
A_3 - ?
- A_1=ArA_1=20 \cdot 0{,}5A_1=10\,\text{см}
We apply the attenuation law specified by the condition.
- A_2=A_1rA_2=10 \cdot 0{,}5A_2=5\,\text{см}
The current amplitude decreases, not the original one every time.
- A_3=A_2rA_3=5 \cdot 0{,}5A_3=2{,}5\,\text{см}
We repeat the calculation for the third period.
Answer2{,}5\,\text{см}
Problem #129
Harmonic, damped and forced oscillationsInitial amplitude 16 cm. After each complete oscillation it is 0{,}75 from the previous one. Find the amplitude after three vibrations.
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- A
- 16\,\text{см}
- r
- 0{,}75\,\text{доля}
A_3 - ?
- A_1=ArA_1=16 \cdot 0{,}75A_1=12\,\text{см}
We apply the attenuation law specified by the condition.
- A_2=A_1rA_2=12 \cdot 0{,}75A_2=9\,\text{см}
The current amplitude decreases, not the original one every time.
- A_3=A_2rA_3=9 \cdot 0{,}75A_3=6{,}75\,\text{см}
We repeat the calculation for the third period.
Answer6{,}75\,\text{см}
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