Photo AI

A student is investigating forced vertical oscillations in springs - AQA - A-Level Physics - Question 7 - 2017 - Paper 1

Question icon

Question 7

A-student-is-investigating-forced-vertical-oscillations-in-springs-AQA-A-Level Physics-Question 7-2017-Paper 1.png

A student is investigating forced vertical oscillations in springs. Two springs, A and B, are suspended from a horizontal metal rod that is attached to a vibration g... show full transcript

Worked Solution & Example Answer:A student is investigating forced vertical oscillations in springs - AQA - A-Level Physics - Question 7 - 2017 - Paper 1

Step 1

1. Deduce the spring constant k for spring A and the mass m suspended from it.

96%

114 rated

Answer

To find the spring constant k, we can use the formula for maximum kinetic energy:

KE=12kA2KE = \frac{1}{2} k A^2

Given that KE=54 mJ=54×103 JKE = 54 \text{ mJ} = 54 \times 10^{-3} \text{ J} and A=2.5×102 mA = 2.5 \times 10^{-2} \text{ m}, we can rearrange to find k:

54×103=12k(2.5×102)254 \times 10^{-3} = \frac{1}{2} k (2.5 \times 10^{-2})^2

Solving for k:

k=2×54×103(2.5×102)2=0.1086.25×104=172.8 N m1k = \frac{2 \times 54 \times 10^{-3}}{(2.5 \times 10^{-2})^2} = \frac{0.108}{6.25 \times 10^{-4}} = 172.8 \text{ N m}^{-1}

For the mass m, using the formula for the angular frequency:

f=12πkmf = \frac{1}{2\pi} \sqrt{\frac{k}{m}}

Rearranging to find m:

m=k(2πf)2m = \frac{k}{(2\pi f)^2}

Substituting k and f = 2.0 Hz:

m=172.8(2π2)2=172.816π20.33 kgm = \frac{172.8}{(2\pi \cdot 2)^2} = \frac{172.8}{16\pi^2} \approx 0.33 \text{ kg}

Step 2

2. Calculate the frequency at which the mass attached to spring B oscillates with maximum amplitude.

99%

104 rated

Answer

Since the stiffness of spring B is 3k, we find:

kB=3×172.8518.4 N m1k_B = 3 \times 172.8 \approx 518.4 \text{ N m}^{-1}

Using the same mass m (0.33 kg), we calculate the frequency:

fB=12πkBm=12π518.40.333.0 Hzf_B = \frac{1}{2\pi} \sqrt{\frac{k_B}{m}} = \frac{1}{2\pi} \sqrt{\frac{518.4}{0.33}} \approx 3.0 \text{ Hz}

Step 3

3. Explain two differences in the graph for spring B.

96%

101 rated

Answer

  1. The peak amplitude for spring B will be at a higher frequency compared to spring A due to the increased stiffness of spring B, which allows for resonant oscillations at a higher frequency.

  2. The width of the resonance peak for spring B will be narrower than that of spring A, indicating less damping effect on oscillations, leading to a sharper peak in the graph.

Join the A-Level students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;